Chiều cao tăng thêm khi thả hòn non bộ:

47 - 35 = 12 (cm) = 0,12 (m)

Thể tích hòn non bộ:

1,2 × 0,4 × 0,12 = 0,576 (m³)

Chọn A

D bạn nhé

B1: bn tính thể tích khi cho hòn non bộ vào

B2: bn tính thế tích khi chưa cho hòn non bộ vào 

B3: lấy cái đầu trừ đi

Bài 2

a) 5/3 - x = 2 1/3

5/3 - x = 7/3

x = 5/3 - 7/3

x = -2/3

b) 3,5 - 1/2 x = -5/4

1/2 x = 3,5 - (-5/4)

1/2 x = 19/4

x = 19/4 : 1/2

x = 19/2

c) 4/(2 - x) - 2/3 = 0

4/(2 - x) = 2/3

2(2 - x) = 3.4

2(2 - x) = 12

2 - x = 12 : 2

2 - x = 6

x = 2 - 6

x = -4

d) 0,25 + 7,5% x = 2 5/6

3/40 x = 17/6 - 0,25

3/40 x = 31/12

x = 31/12 : 3/40

x = 310/9

Bài 4

a) Số học sinh xếp loại tốt:

120 . 4/15 = 32 (học sinh)

Số học sinh xếp loại khá:

32 : 80% = 40 (học sinh)

Số học sinh xếp loại đạt:

120 - 32 - 40 = 48 (học sinh)

b) Tỉ số phần trăm của số học sinh xếp loại khá so với cả khối:

40 . 100% : 120 ≈ 33,33%

Điều kiện:

 \(\left\{{}\begin{matrix}x+\dfrac{3}{x}=\dfrac{x^2+3}{x}\ge0\\\dfrac{x^2+7}{2\left(x+1\right)}\ge0\end{matrix}\right.\)

mà \(x^2\ge0\forall x\Rightarrow\left\{{}\begin{matrix}x^2+3>0\forall x\\x^2+7>0\forall x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2+3}{x}\ge0\\\dfrac{x^2+7}{2\left(x+1\right)}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\2\left(x+1\right)>0\Leftrightarrow x+1>0\Leftrightarrow x>-1\end{matrix}\right.\)

\(\Leftrightarrow x>0\)

\(\sqrt{x+\dfrac{3}{x}}=\dfrac{x^2+7}{2\left(x+1\right)}\)

\(\Leftrightarrow\sqrt{\dfrac{x^2+3}{x}}=\dfrac{x^2+7}{2\left(x+1\right)}\)

\(\Leftrightarrow\left(\sqrt{\dfrac{x^2+3}{x}}\right)^2=\left[\dfrac{x^2+7}{2\left(x+1\right)}\right]^2\)

\(\Leftrightarrow\dfrac{x^2+3}{x}=\dfrac{\left(x^2+7\right)^2}{\left[2\left(x+1\right)\right]^2}\)

\(\Leftrightarrow\dfrac{x^2+3}{x}=\dfrac{x^4+14x^2+49}{4\left(x+1\right)^2}=\dfrac{x^4+14x^2+49}{4\left(x^2+2x+1\right)}=\dfrac{x^4+14x^2+49}{4x^2+8x+4}\)

\(\Leftrightarrow\dfrac{\left(x^2+3\right)\left(4x^2+8x+4\right)}{x\left(4x^2+8x+4\right)}=\dfrac{x\left(x^4+14x^2+49\right)}{x\left(4x^2+8x+4\right)}\)

\(\Leftrightarrow\left(x^2+3\right)\left(4x^2+8x+4\right)=x\left(x^4+14x^2+49\right)\)

\(\Leftrightarrow x^2\left(4x^2+8x+4\right)+3\left(4x^2+8x+4\right)=x\left(x^4+14x^2+49\right)\)

\(\Leftrightarrow4x^4+8x^3+4x^2+12x^2+24x+12=x^5+14x^3+49x\)

\(\Leftrightarrow4x^4+8x^3+16x^2+24x+12=x^5+14x^3+49x\)

\(\Leftrightarrow x^5-4x^4+14x^3-8x^3-16x^2+49x-24x-12=0\)

\(\Leftrightarrow x^5-4x^4+6x^3-16x^2+25x-12=0\)

\(\Leftrightarrow x^5-x^4-3x^4+3x^3+3x^3-3x^2-13x^2+13x+12x-12=0\)

\(\Leftrightarrow x^4\left(x-1\right)-3x^3\left(x-1\right)+3x^2\left(x-1\right)-13x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^4-3x^3+3x^2-13x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^4-x^3-2x^3+2x^2+x^2-x-12x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^3\left(x-1\right)-2x^2\left(x-1\right)+x\left(x-1\right)-12\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^3-2x^2+x-12\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^3-2x^2+x-12\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^3-3x^2+x^2-3x+4x-12\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^2\left(x-3\right)+x\left(x-3\right)+4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-3\right)\left(x^2+x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\\x^2+x+4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(tm\right)\\x^2+x+\dfrac{1}{4}+\dfrac{15}{4}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(tm\right)\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\end{matrix}\right.\)

Có: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\forall x\)

\(\Rightarrow x^2+x+4=0\) vô nghiệm

Vậy: \(x\in\left\{1;3\right\}\)

kinh thật!

mn giúp mình nha

Bài giải:

Tổng số tấn thóc của 2 kho là:

\(105\) x \(2=210\) (tấn)

Số tấn thóc ở kho thứ hai là:

\(\left(210+20\right):2=115\) (tấn)

Số tấn thóc ở kho thứ nhất là:

\(210-115=95\) (tấn)

Đáp số : \(95\) tấn

Bài 5:

1: \(\dfrac{2}{5}+\dfrac{3}{5}x=\dfrac{9}{20}\)

=>\(\dfrac{3}{5}x=\dfrac{9}{20}-\dfrac{2}{5}=\dfrac{9}{20}-\dfrac{8}{10}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{5}{30}=\dfrac{1}{6}\)

2: \(\dfrac{1}{4}+\dfrac{1}{3}:\left(3x\right)=-5\)

=>\(\dfrac{1}{3}:\left(3x\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

=>\(3x=-\dfrac{1}{3}:\dfrac{21}{4}=-\dfrac{1}{3}\cdot\dfrac{4}{21}=-\dfrac{4}{63}\)

=>\(x=-\dfrac{4}{63}:3=-\dfrac{4}{189}\)

3: \(-\dfrac{21}{13}x+\dfrac{1}{3}=-\dfrac{2}{3}\)

=>\(-\dfrac{21}{13}x=-\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{3}{3}=-1\)

=>\(x=1:\dfrac{21}{13}=\dfrac{13}{21}\)

4: \(\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\)

=>\(\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{3}{7}=\dfrac{7+6}{14}=\dfrac{13}{14}\)

=>\(x=\dfrac{13}{14}:\dfrac{3}{4}=\dfrac{13}{14}\cdot\dfrac{4}{3}=\dfrac{52}{42}=\dfrac{26}{21}\)

5: \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{3}{10}-\dfrac{1}{5}\)

=>\(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)

=>\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}=\dfrac{1}{10}-\dfrac{5}{10}=-\dfrac{4}{10}=-\dfrac{2}{5}\)

=>\(x=-\dfrac{2}{5}:\dfrac{2}{3}=-\dfrac{2}{5}\cdot\dfrac{3}{2}=-\dfrac{3}{5}\)

6: \(\dfrac{2}{3}+\dfrac{1}{3}x=-\dfrac{1}{2}\)

=>\(\dfrac{1}{3}x=-\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{-7}{6}\)

=>\(x=-\dfrac{7}{6}:\dfrac{1}{3}=-\dfrac{7}{6}\cdot3=-\dfrac{7}{2}\)

7: \(\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)

=>\(\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}=\dfrac{-2}{12}-\dfrac{9}{12}=-\dfrac{11}{12}\)

=>x=-1

8: \(-\dfrac{5}{6}-\dfrac{2}{3}x=\dfrac{7}{12}+\dfrac{-1}{3}\)

=>\(-\dfrac{5}{6}-\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{4}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

=>\(\dfrac{2}{3}x=-\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{-10}{12}-\dfrac{3}{12}=-\dfrac{13}{12}\)

=>\(x=-\dfrac{13}{12}:\dfrac{2}{3}=-\dfrac{13}{12}\cdot\dfrac{3}{2}=\dfrac{-39}{24}=\dfrac{-13}{8}\)

9: \(2\dfrac{2}{3}x+8\dfrac{2}{3}=3\dfrac{1}{3}\)

=>\(\dfrac{8}{3}x=3+\dfrac{1}{3}-8-\dfrac{2}{3}=-5-\dfrac{1}{3}=-\dfrac{16}{3}\)

=>\(x=-\dfrac{16}{3}:\dfrac{8}{3}=-\dfrac{16}{3}\cdot\dfrac{3}{8}=-\dfrac{16}{8}=-2\)

10: \(\dfrac{3}{4}-\dfrac{1}{4}:\left(x:\dfrac{5}{2}\right)=-3\)

=>\(\dfrac{1}{4}:\left(x:\dfrac{5}{2}\right)=\dfrac{3}{4}+3=\dfrac{15}{4}\)

=>\(x:\dfrac{5}{2}=\dfrac{1}{4}:\dfrac{15}{4}=\dfrac{1}{15}\)

=>\(x=\dfrac{1}{15}\cdot\dfrac{5}{2}=\dfrac{1}{6}\)

B.4

B. 4.

Bài 1:

a: \(A\left(x\right)=2x^4+3x^2-x+3-x^2-x^4-6x^3\)

\(=\left(2x^4-x^4\right)-6x^3+\left(3x^2-x^2\right)-x+3\)

\(=x^4-6x^3+2x^2-x+3\)

\(B\left(x\right)=10x^3+3-x^4-4x^3+4x-2x^2\)

\(=-x^4+\left(10x^3-4x^3\right)-2x^2+4x+3\)

\(=-x^4+6x^3-2x^2+4x+3\)

b: M(x)=A(x)+B(x)

\(=x^4-6x^3+2x^2-x+3-x^4+6x^3-2x^2+4x+3\)

=3x+6

N(x)=A(x)-B(x)

\(=x^4-6x^3+2x^2-x+3+x^4-6x^3+2x^2-4x-3\)

\(=2x^4-12x^3+4x^2-5x\)

c: M(x)=0

=>3x+6=0

=>3x=-6

=>x=-2

Bài 3: 

a: \(A\left(x\right)=x-5x^3-2x^2+9x^3-\left(x-1\right)-2x^2\)

\(=\left(-5x^3+9x^3\right)+\left(-2x^2-2x^2\right)+\left(x-x\right)+1\)

\(=4x^3-4x^2+1\)

\(B\left(x\right)=-4x^3-2\left(x^2+1\right)+6x+2x^2-9x+2x^3\)

\(=\left(-4x^3+2x^3\right)+\left(-2x^2+2x^2\right)+6x-9x-2\)

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

\(C\left(x\right)=2x-6x^2-4+x^3\)

\(=x^3-6x^2+2x-4\)

b: M(x)=A(x)+B(x)-C(x)

\(=4x^3-4x^2+1-2x^3-3x-2-x^3+6x^2-2x+4\)

\(=x^3+2x^2-5x+3\)

c: \(P\left(x\right)=3\cdot M\left(x\right)-3x^3-9\)

\(=3x^3+6x^2-15x+9-3x^3-9=6x^2-15x\)

Đặt P(x)=0

=>\(6x^2-15x=0\)

=>\(2x^2-5x=0\)

=>x(2x-5)=0

=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

1962

Truyện ngắn Pauxtốpxki, Hà Nội, 1962.

C.will be talking

A. Will talk.