a: \(\Delta=\left[2\left(m+1\right)\right]^2-4\cdot2\cdot m\)

\(=\left(2m+2\right)^2-8m=4m^2+8m+4-8m=4m^2+4>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

b: Theo Vi-et, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m+1\right)}{2}=-\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=\dfrac{m}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1< 1\\x_2< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2< 2\\x_1\cdot x_2< 1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\left(m+1\right)< 2\\\dfrac{m}{2}< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>-2\\m< 2\end{matrix}\right.\)

=>-3<m<2

a) 

\(\Delta=\left[2\left(m+1\right)\right]^2-4\cdot2\cdot m=4\left(m^2+2m+1\right)-8m\\ =4m^2+8m+4-8m=4m^2+4\ge4>0\forall x\)

=> Pt luôn có nghiệm với mọi m 

b) Ta có: \(\left\{{}\begin{matrix}x_1< 1\\x_2< 1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x_1+x_2< 2\\x_1x_2< 1\end{matrix}\right.\) 

Theo vi-ét: \(\left\{{}\begin{matrix}x_1x_2=\dfrac{m}{2}\\x_1+x_2=\dfrac{-2\left(m+1\right)}{2}=-\left(m+1\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{m}{2}< 1\\-\left(m+1\right)< 2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 2\\m>-3\end{matrix}\right.\Rightarrow-3< m< 2\)

Số số hạng trong tích này là:

(2024-34):10+1=200(số)

Vì 200 chia hết cho 4

nên \(4^{200}\) có chữ số tận cùng là 6

\(34\times44\times...\times2024\) sẽ có chữ số tận cùng giống với chữ số tận cùng của 4x4x4x...4x4(200 chữ số 4)

=>34x44x...x2024 có chữ số tận cùng là 6

a: \(\left[\left(-21,8\right)+4,125\right]+\left[11,8+\left(-2,125\right)\right]\)

=-21,8+4,125+11,8-2,125

=(4,125-2,125)+(-21,8+11,8)

=2-10

=-8

b: \(\left(-124,5\right)+\left(-6,24+124,5\right)\)

\(=-124,5-6,24+124,5\)

=-6,24

\(a,\left[\left(-21,8\right)+4,125\right]+\left[11,8+\left(-2,125\right)\right]\)

\(=-21,8+4,125+11,8-2,125\)

\(=\left[\left(-21,8\right)+11,8\right]+\left(4,125-2,125\right)\)

\(=-10+2\)

\(=-8\)

\(b,\left(-124,5\right)+\left(-6,24+124,5\right)\)

\(=-124,5-6,24+124,5\)

\(=\left[\left(-124,5\right)+124,5\right]-6,24\)

\(=0-6,24\)

\(=-6,24\)

$\color{#90EE90}{\text{4}}$  $\color{#B0E0E6}{\text{56}}$

a: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-y-3x+y=4-5\\3x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\y=3x-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3x-5=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=-3+9=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=y\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0=2\left(vôlý\right)\\x=y\end{matrix}\right.\)

vậy: Hệ vô nghiệm

a) 

\(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3x-y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\-\dfrac{3}{2}-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-y=2\end{matrix}\right.\)

mà: 2 khác 0

=> Hpt vô nghiệm 

Số phần quyển sách còn lại sau ngày thứ nhất là:

\(1-40\%=60\%=\dfrac{3}{5}\)

Số phần quyển sách còn lại sau ngày thứ hai là:

\(\dfrac{3}{5}\times\left(1-60\%\right)=\dfrac{3}{5}\times\dfrac{2}{5}=\dfrac{6}{25}\)

Số phần quyển sách còn lại sau ngày thứ ba là:

\(\dfrac{6}{25}\times\left(1-80\%\right)=\dfrac{6}{25}\times\dfrac{1}{5}=\dfrac{6}{125}\)

Số trang của quyển sách là:\(30:\dfrac{6}{125}=30\times\dfrac{125}{6}=625\left(trang\right)\)

giúp ik mik tích choa =))

Cậu chia nhỏ bài ra để được hỗ trợ nhanh hơn nhé!

$\color{#0000CD}{\text{A}}$   $\color{#0000FF}{\text{n}}$
$\color{#8A2BE2}{\text{nn}}$

a: ĐKXĐ: \(x\ne0;y\ne0\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Hệ phương trình sẽ trở thành: \(\left\{{}\begin{matrix}a-2b=-1\\2a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-2b=-1\\4a+2b=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a-2b+4a+2b=-1+6\\2a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=5\\b=3-2a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=1\\b=3-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=1\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(nhận\right)\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne-\dfrac{y}{2}\end{matrix}\right.\)

Đặt \(\dfrac{1}{x-y}=a;\dfrac{1}{2x+y}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}a+b=2\\3a-2b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+2b=4\\3a-2b=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2a+2b+3a-2b=4-2\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=2\\b=2-a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\dfrac{2}{5}\\b=2-\dfrac{2}{5}=\dfrac{10}{5}-\dfrac{2}{5}=\dfrac{8}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x-y}=\dfrac{2}{5}\\\dfrac{1}{2x+y}=\dfrac{8}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{5}{2}\\2x+y=\dfrac{5}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y+2x+y=\dfrac{5}{2}+\dfrac{5}{8}\\x-y=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{25}{8}\\y=x-\dfrac{5}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{25}{8}:3=\dfrac{25}{24}\\y=\dfrac{25}{24}-\dfrac{5}{2}=\dfrac{25}{24}-\dfrac{60}{24}=-\dfrac{35}{24}\end{matrix}\right.\left(nhận\right)\)

c: ĐKXĐ: \(x\ne1;y\ne-3\)

Đặt \(\dfrac{x}{x-1}=a;\dfrac{1}{y+3}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}3a-2b=3\\4a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=3\\8a+2b=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3a-2b+8a+2b=13\\4a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11a=13\\b=5-4a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{13}{11}\\b=5-4\cdot\dfrac{13}{11}=\dfrac{55}{11}-\dfrac{52}{11}=\dfrac{3}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=\dfrac{13}{11}\\\dfrac{1}{y+3}=\dfrac{3}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13\left(x-1\right)=11x\\y+3=\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x-13=11x\\y=\dfrac{11}{3}-3=\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{13}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\left(nhận\right)\)

a) 

\(\left(-12,5\right)+3.4+12,5+\left(-3,4\right)\\ =\left(12,5-12,5\right)+\left(3,4-3,4\right)\\ =0+0=0\)

b) 

\(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\\ =\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\\ =0+0+4,2\\ =4,2\)

c) 

\(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\\ =2,5\left(-42,5-150-7,5\right)\\ =2,5\cdot\left(-50-150\right)\\ =2,5\cdot-200\\ =-500\) 

d) 

\(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\\ =2,6\cdot\left(-2,45-7,55\right)\\ =2,6\cdot-10\\ =-26\)

a: \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)\)

\(=\left(-12,5+12,5\right)+\left(3,4-3,4\right)\)

=0+0=0

b: \(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\)

\(=\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\)

=0+0+4,2

=4,2

c: \(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\)

\(=2,5\cdot\left(-42,5-150-7,5\right)\)

\(=2,5\cdot\left(-200\right)=-500\)

d: Sửa đề: \(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\)

\(=2,6\left(-2,45-7,55\right)\)

\(=2,6\cdot\left(-10\right)=-26\)

2:

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)

c: \(x^2+12x+36=x^2+2\cdot x\cdot6+6^2=\left(x+6\right)^2\)

d: \(4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\)

e: \(9x^2+6x+1=\left(3x\right)^2+2\cdot3x\cdot1+1^2=\left(3x+1\right)^2\)

f: \(16x^2+24x+9=\left(4x\right)^2+2\cdot4x\cdot3+3^2=\left(4x+3\right)^2\)

3:

a: \(A=\left(x+2\right)^2-x\left(x+3\right)+4x-3\)

\(=x^2+4x+4-x^2-3x+4x-3\)

=5x+1

b: \(B=\left(x+3\right)^2-x\left(x-5\right)+7x-8\)

\(=x^2+6x+9-x^2+5x+7x-8\)

=18x+1

c: \(C=\left(2x+3\right)^2-x\left(x+4\right)-9x-3\)

\(=4x^2+12x+9-x^2-4x-9x-3\)

\(=3x^2-x+6\)

d: \(D=\left(2x+21\right)^2-2x\left(2x-4\right)-5x-21\)

\(=4x^2+84x+441-4x^2+8x-5x-21\)

=87x+420

2:

\(a.x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\\ b.x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\\ c.x^2+12x+36=x^2+2\cdot x\cdot6+6^2=\left(x+6\right)^2\\ d.4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\\ e.9x^2+6x+1=\left(3x\right)^2+2\cdot3x\cdot1+1^2=\left(3x+1\right)^2\\ f.16x^2+24x+9=\left(4x\right)^2+2\cdot4x\cdot3+3^2=\left(4x+3\right)^2\)