1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)

\(\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

     \(\text{Vậy x=-5}\)

\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)

\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)

\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)

\(\Rightarrow-16x-8=7\)

\(\Rightarrow-16x=15\)

\(\Rightarrow x=\frac{-15}{16}\)

      \(\text{Vậy }x=\frac{-15}{16}\)

\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)

\(\Rightarrow-9+8x-1=8\)

\(\Rightarrow8x=18\)

\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)

      \(\text{Vậy }x=\frac{9}{4}\)

\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)

a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)

 PT <=> 2x - 1 = 5

<=> x = 3 ( TM )

Vậy ...

b, ĐKXĐ : \(x\ge5\)

PT <=> x - 5 = 9

<=> x = 14 ( TM )

Vậy ...

c, PT <=> \(\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy ...

d, PT<=> \(\left|x-3\right|=3-x\)

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

Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)

e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)

PT <=> 2x + 5 = 1 - x

<=> 3x = -4

<=> \(x=-\dfrac{4}{3}\left(TM\right)\)

Vậy ...

f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

PT <=> \(x^2-x=3-x\)

\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )

Vậy ...

a) \(\sqrt{2x-1}=\sqrt{5}\)          (x \(\ge\dfrac{1}{2}\))

<=> 2x - 1 = 5

<=> x = 3 (tmđk)

Vậy S = \(\left\{3\right\}\)

b) \(\sqrt{x-5}=3\)           (x\(\ge5\))

<=> x - 5 = 9

<=> x = 4 (ko tmđk)

Vậy x \(\in\varnothing\)

c) \(\sqrt{4x^2+4x+1}=6\)          (x \(\in R\))

<=> \(\sqrt{\left(2x+1\right)^2}=6\)

<=> |2x + 1| = 6

<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)

Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)

a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)

\(\Leftrightarrow x=25\)

b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x=1\)

hay \(x=-\dfrac{1}{9}\)

a: ta có: $\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15$

$\iff 2{}^{}+4x-5x-10-2{}^{}+2x=15$

$\iff x=25$

b: Ta có: $\left(5-2x\right)\left(2x+7\right)=4{}^{}-25$

$\iff 4{}^{}-25+\left(2x-5\right)\left(2x+7\right)=0$

$\iff \left(2x-5\right)(2x+5+2x+$