\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)

\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)

\(\text{⇔}8x+5=-11\) 

\(\text{⇔}8x=-16\)

\(\text{⇔}x=-2\)

Vậy: \(x=-2\)

==========

\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)

\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x=-\dfrac{4}{3}\)

\(\text{⇔}x=\dfrac{2}{39}\)

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

\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)

\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)

\(x\cdot\frac{19}{6}=-\frac{1}{24}\)

x = -1/76

b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

\(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

2x = 12

x = 6

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

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

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

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

x(2x - 3) - 2(3 - 2x) = 0

x(2x - 3) + 2(2x - 3) = 0

(2x - 3)(x + 2) = 0

\(\left[\begin{array}{nghiempt}2x-3=0\\x+2=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-2\end{array}\right.\)

2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = - 26 : 13

x = - 2

Bài 2: 

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

=>-13x=26

hay x=-2

b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)

c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

hay \(x\in\left\{-5;2\right\}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`(2x - 1)^2 + 1 = 26`

`\Rightarrow (2x - 1)^2 = 26 - 1`

`\Rightarrow (2x - 1)^2 = 25`

`\Rightarrow (2x - 1)^2 = (+-5)^2`

`\Rightarrow`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=6\div2\\x=-4\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-2;3\right\}\)

`b)`

`(2x - 4)^3 + 2 = 66`

`\Rightarrow (2x - 4)^3 = 66 - 2`

`\Rightarrow (2x - 4)^3 = 64`

`\Rightarrow (2x - 4)^3 = 4^3`

`\Rightarrow 2x - 4 = 4`

`\Rightarrow 2x = 8`

`\Rightarrow x = 8 \div 2`

`\Rightarrow x = 3`

Vậy, `x = 3`

`c)`

\(7^{x+2}+5\cdot7^{x+1}+15=603\)

`\Rightarrow 7^x . 7^2 + 5 . 7^x . 7 = 603 - 15`

`\Rightarrow 7^x . 7^2 + 35 . 7^x = 588`

`\Rightarrow 7^x . (7^2 + 35) = 588`

`\Rightarrow 7^x . 84 = 588`

`\Rightarrow 7^x = 588 \div 84`

`\Rightarrow 7^x = 7`

`\Rightarrow 7^x = 7^1`

`\Rightarrow x = 1`

Vậy, `x = 1.`

\(#48Cd\)

\(a)\left(2x-1\right)^2+1=26\)

\(\left(2x-1\right)^2=25\)

\(TH1:2x-1=5\)

\(2x=6\)

\(x=3\)

\(TH2:2x-1=-5\)

\(2x=-4\)

\(x=-2\)

Vậy........

\(b)\left(2x-4\right)^3+2=66\)

\(\left(2x-4\right)^3=64=4^3\)

\(2x-4=4\)

\(2x=8\)

\(x=4\)

\(c)7^x+2+5.7^x+1+15=603\)

\(7^x\left(1+5\right)=603-15-1-2\)

\(7^x.6=585\)

Bạn xem lại phần này nhé . x tìm ra không được chẵn lắm á cậu.

Bạn xem lại câu a

a,(2x-1)²+1=26

=>(2x-1)²=26-1=25=5²

=>2x-1=5 và -5

=>x=(5+1):2=3 và x=(-5+1):2=-2

b,(2x-4)³+2=66

=>(2x-4)³=66-2=64=4³

=>2x-4=4

=>x=(4+4):2=4

c,7^x+2+5\(\cdot\)7^x+1+15=603

=>7^x\(\cdot\)7²+5\(\cdot\)7^x\(\cdot\)7=603-15=588

=>7^x(7²+5\(\cdot\)7)=588

=>7^x\(\cdot\)84=588

=>7^x=588:84=7

=>x=1

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

hay x=-2