a. x=0

b.x=0

cau c 

tk ung ho mk nha

\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)

a ) Ta có : - 12 . ( x - 5 ) + 7( 3 - x ) = 5

Suy ra - 12x - ( - 12 ) . 5 + 7 . 3 - 7x = 5

Suy ra - 12x + 60 + 21 - 7x = 5

Suy ra - 12x - 7x = 5 - 60 - 21

Suy ra - 19x = - 76

Suy ra x = -76 : ( - 19 )

Vậy x = 4

b ) Ta có : 30 . ( x + 2 ) - 6  . ( x - 5 ) - 24x = 100

Suy ra 30x + 30 . 2 - 6x - ( - 6 ) . 5 - 24x = 100

Suy ra 30x + 60 - 6x + 30 - 24x = 100

Suy ra 30x - 6x - 24x = 100 - 60 - 30

Suy ra 0x = 10

Vậy x = 0

a: \(\Leftrightarrow x\left(16-x^2\right)+x^3-125=3\)

=>16x-125=3

=>16x=128

hay x=8

b: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

=>12x-4=-10

=>12x=-6

hay x=-1/2

c: \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

\(\Leftrightarrow4x-27=1\)

hay x=7

a/ | x + 10 | = 15

=> x + 10 = 15 hay x + 10 = - 15

+/ x + 10 = 15

=> x = 15 - 10 = 5

+/ x + 10 = - 15

=> x = -15 - 10 = -25

Vậy x thuộc {5; - 25}

b/ | x - 3 | + 5 = 7

=> | x - 3 | = 7 - 5 = 2

=> x - 3 = 2 hay x - 3 = -2

+/ x - 3 = 2

=> x = 2+3 = 5

+/ x - 3 = -2

=> x = -2 + 3 = 1

Vậy x thuộc {5;1}

c/ | x - 3 | + 12 = 6

=> | x - 3 | = 6 - 12 = - 6

Vì | x - 3 | luôn > 0

mà | x - 3 | = - 6

Vậy k có giá trị của x

d/ (2x + 4) . (3x + 9) = 0

=> 2x + 4 = 0 hoặc 3x + 9 = 0

+/ 2x + 4 = 0

=> 2x = 0 - 4 = -4

=> x = (-4) / 2 = -2

+/ 3x - 9 = 0

=> 3x = 0 + 9 = 9

=> x = 9 / 3 = 3

Vậy x thuộc {-2;3}

a. \(\left|x+10\right|=15\)

\(\Rightarrow x+10=\pm15\)

\(TH1:x+10=15\)

\(x=15-10\)=5

TH2: x + 10 = -15

x = -15 -10 = -25

Vậy x \(\in\left\{5;-25\right\}\)

b. \(\left|x-3\right|+5=7\)

\(\left|x-3\right|=7-5=2\)

\(\Rightarrow x-3=\pm2\)

TH1: x - 3 = 2

x = 2 + 3 = 5

TH2: x - 3 = -2

x = -2 + 3 = 1

Vậy x \(\in\left\{5;-1\right\}\)

* Đối với bài tập về phép đối này thì có 2 trường hợp, giải TH âm và dương của số đã cho bên kết quả.

Mỏi tay, xl

bài e đề là x² - 5x = 0 ak