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4 tháng 2 2020

a) Đặt x -3 = a

<=> a(a+2)(a+8)(a+10) - 297=0

<=> \(\left[a\left(a+10\right)\right]\left[\left(a+2\right)\left(a+8\right)\right]\)-297=0

<=> \(\left(a^2+10a\right)\left(a^2+10a+16\right)-297=0\)

Đặt \(a^2+10a=b\)

\(b^2+16b-297=0\)

\(\Rightarrow\left[{}\begin{matrix}b=11\\b=-27\end{matrix}\right.\)\(b=11\Rightarrow\left[{}\begin{matrix}a=1\\a=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

b= -27 \(\Rightarrow a=\varnothing\Rightarrow x=\varnothing\)

4 tháng 2 2020

b) bấm máy ra nhân tử chung :D

c)

\(\Leftrightarrow\left(\frac{1927-X}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+...=0\)

\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

<=> x = 2018

d) \(\Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-3\right)=0\)

giống câu c

a: \(\Leftrightarrow x+2016=0\)

hay x=-2016

b: \(\Leftrightarrow x-100=0\)

hay x=100

11 tháng 3 2021

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

17 tháng 7

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

\(b^2\) + 2b + 1 = 25

\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

28 tháng 12 2017

4.

\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)

29 tháng 12 2017

\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)

\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)

\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)

AH
Akai Haruma
Giáo viên
5 tháng 1 2019

Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)

\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)

\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)

AH
Akai Haruma
Giáo viên
5 tháng 1 2019

Bài 2:

\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)

\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)

5 tháng 1 2018

Phương trình đưa được về dạng ax + b = 0Phương trình đưa được về dạng ax + b = 0

9 tháng 1 2019

Câu a bạn có thể làm cho mình được ko. Mk đang cần nó

4 tháng 1 2019

Bạn vất vả r

11 tháng 3 2023

bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu

a: =>\(4x-5=2x-2+x=3x-2\)

=>x=3

b: \(\Leftrightarrow7x-35=3x+6\)

=>4x=41

=>x=41/4

c: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25-2x^2=0

=>15x=-25

=>x=-5/3

e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)

=>11(x^2-3x-4)=x(11x-34)

=>11x^2-33x-44=11x^2-34x

=>x=44

13 tháng 1 2018

bài 1:

\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)

<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)

<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)

vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0

nên x-2004=0=>x=2004

vyaj.......

bài 2:

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)

<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)

<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)

vì 1/15+1/13+1/11+1/9 khác 0

=>x-100=0<=>x=100

4 tháng 2 2018

a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)

<=> \(25x+10-80x+10=24x+12-30\)

<=> \(25x-80x-24x=12-30-10-10\)

<=> \(-79x=-38\)

<=> \(x=\dfrac{-38}{-79}\)

\(x=\dfrac{38}{79}\)

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

<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)

<=> \(30x-12x+30+5x+40=210+10x-10\)

<=> \(30x-12x+5x-10x=210-10-30-40\)

<=> \(13x=130\)

<=> \(x=\dfrac{130}{13}\)

\(x=10\)

c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)

<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)

<=> \(28x+28+60x+120+105x+420+2520=0\)

<=> \(28x+60x+105x=-28-120-420-2520\)

<=> \(193x=-3088\)

<=> \(x=\dfrac{-3088}{193}\)

\(x=-16\)

d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)

<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)

<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)

<=> \(22968x=8199576\)

<=> \(x=\dfrac{8199576}{22968}\)

\(x=357\)

4 tháng 2 2018

Đề là giải PT nha các bn