\(\left|x+1\right|-\left|-2x-2\right|=2\)

\(\Leftrightarrow\left|x+1\right|-\left|-2\left(x+1\right)\right|=2\)

\(\Leftrightarrow\left|x+1\right|-2\left|x+1\right|=2\)

\(\Leftrightarrow-\left|x+1\right|=2\)

\(\Leftrightarrow\left|x+1\right|=-2\)

\(\Leftrightarrow\left|x+1\right|+2=0\)

Mà: \(\left|x+1\right|\ge0\forall x\Rightarrow\left|x+1\right|+2\ge2>0\)

\(\Leftrightarrow\left|x+1\right|+2=0\) (vô lí)

Vậy phương trình vô nghiệm:

\(x\in\varnothing\)

=>|x+1|-2|x+1|=2

=>-|x+1|=2

=>|x+1|=-2(vô lý)

Vậy: \(x\in\varnothing\)

\(TH_1:\left\{{}\begin{matrix}x+1\ge0\\-2x-2\ge0\end{matrix}\right.\) 

\(\left(1\right)\left(x+1\right)-\left(-2x-2\right)=2\) 

\(\Leftrightarrow x+1+2x+2=2\)

\(\Leftrightarrow3x+3=2\)  

\(\Leftrightarrow x=-\dfrac{1}{3}\left(l\right)\)

\(TH_2:\left\{{}\begin{matrix}x+1< 0\\-2x-2< 0\end{matrix}\right.\) 

\(\left(1\right)\left(x+1\right)-\left(-\left(-2x-2\right)\right)=2\) 

\(\Leftrightarrow-x-1+2x+2=2\) 

\(\Leftrightarrow x+1=2\)

\(\Leftrightarrow x=1\left(l\right)\) 

Vậy tập nghiệm rỗng.

a) Ta có: 2x+33=-11

nên 2x=-44

hay x=-22

b) Ta có: \(\dfrac{x}{2}=\dfrac{-49}{14}\)

nên x=-7

c) Ta có: \(\dfrac{5}{6}x+\dfrac{10}{3}=\dfrac{7}{2}\)

nên \(\dfrac{5}{6}x=\dfrac{7}{2}-\dfrac{10}{3}=\dfrac{1}{6}\)

hay \(x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\)

1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

=>2x^2+2x-3x-3+x^2+2x=3x^2+12x+12

=>12x+12=x-3

=>11x=-15

=>x=-15/11

b, ( 5/2 - x ) ^2

=25/4-4/5x+x^2

c,( xy/2 - x/3 ) ( xy/2 + x/3)

=(xy/2)^2-(x/3)^2

c: \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)

e: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

b) \(\left(\dfrac{5}{2}-x\right)^2=\dfrac{25}{4}-5x+x^2\)

c) \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)

d) \(\left(3x+\dfrac{2}{3}yz\right)^2=9x^2+4xyz+\dfrac{4}{9}y^2z^2\)

e) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

f) \(\left(2x-y+2\right)^2=\left(2x-y\right)^2+4\left(2x-y\right)+2^2=4x^2+y^2+4-4xy+8x-4y\)

TRẢ LỜI:

2x2-3x-2x2-4

ĐKXĐ: x ≠ 2 hoặc x ≠ -2

⇔ 2x2-3x-2=2x2-4 ⇔ 2x2-3x-2=2x2-8

⇔ 2x2-2x2-3x = - 8 + 2 ⇔ - 3x = - 6 ⇔ x = 2 (loại)

Vậy không có giá trị nào của x thỏa mãn điều kiện bài toán.

sai thui!

\(\Rightarrow\dfrac{2}{3}:x=\dfrac{5}{3}\Rightarrow x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

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

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)