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

\(x=\dfrac{1}{10}\)

\(b,x+\dfrac{3}{7}=\dfrac{7}{10}\)

\(x=\dfrac{7}{10}-\dfrac{3}{7}\)

\(x=\dfrac{19}{70}\)

\(c,19-x=\dfrac{17}{20}\)

\(x=19-\dfrac{17}{20}\)

\(x=\dfrac{363}{20}\)

a,1/10

b,19/70

c,18/4

x=36320

a: =>x:2,2=15,2x2=30,4

=>x=66,88

b: =>(15/4x+3/4)=-2/3

=>15/4x=-17/12

hay x=-17/45

a) |x| = 2,5

=>\(\left[\begin{array}{nghiempt}x=2,5\\x=-2,5\end{array}\right.\)

vậy x=2,5 hoặc x=-2,5

b)|x|=-1,2

=>x không có giá trị thỏa mãn |x|\(\ge\) 0

c)|x| + 0,573 = 2

|x| = 2 - 0,573

|x| = 1,427

=>\(\left[\begin{array}{nghiempt}x=1,427\\x=-1,427\end{array}\right.\)

Vậy x = 1,427 hoặc x = -1,427

d) ∣∣x+13∣∣ - 4 = -1

=>|x+\(\frac{1}{3}\)| =-1 + 4

|x+\(\frac{1}{3}\)| = 3

.....................

Vậy x = \(\frac{8}{3}\) hoặc x = \(\frac{-10}{3}\)

a ) \(\left|x\right|=2,5\Rightarrow x=2,5;x=-2,5\)

b ) \(\left|x\right|=-1,2\Rightarrow\left|x\right|\ge0\forall x\Rightarrow x\in\varnothing\)

c ) \(\left|x\right|+0,573=2\)

\(\Rightarrow\)\(\left[\begin{array}{nghiempt}x+0,573=2\\x+0,573=-2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}2-0,573\\\left(-2\right)-0,573\end{array}\right.\) \(\Rightarrow\)\(\left[\begin{array}{nghiempt}x=1,427\\x=-2,573\end{array}\right.\)

Vậy \(x\in1,427;-2,573\)

d ) \(\left|x+\frac{1}{3}\right|-4=-1\)

\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=3-\frac{1}{3}\\x=\left(-3\right)-\frac{1}{3}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=\frac{-10}{3}\end{array}\right.\)

Vậy \(x\in\frac{8}{3};\frac{-10}{3}\)

(Gíup mình với các thần đồng 's hoc24 ơi)

a) \(\left|x\right|=2,5\Rightarrow\)\(\left[{}\begin{matrix}x=2,5\\x=-2,5\end{matrix}\right.\)

b) \(\left|x\right|=-1,2\left(VLý\right)\Rightarrow S=\varnothing\)

c) \(\left|x\right|+0,573=2\Rightarrow\left|x\right|=1,427\)

\(\Rightarrow\left[{}\begin{matrix}x=1,427\\x=-1,427\end{matrix}\right.\)

d) \(\left|x+\dfrac{1}{3}\right|-4=-1\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{100}{3}\end{matrix}\right.\)

Câu a) -3 phần 1/2 

Câu a) 2 mũ 2

a: =>-7/2:(4/5-1/2x)=4

=>4/5-1/2x=-7/2:4=-7/8

=>1/2x=4/5+7/8=67/40

=>x=67/20

b: =>5x=5

=>x=1

c: =>x(-2/3-1/3)=-2

=>-x=-2

=>x=2

d: =>-2/3(x+1)=-1/3+1/2=1/6

=>x+1=-1/6:2/3=-1/6*3/2=-3/12=-1/4

=>x=-1/4-1=-5/4

a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)

hay \(x\in\left\{30;-30\right\}\)

Vậy: \(x\in\left\{30;-30\right\}\)

b) Ta có: \(\left|x\right|+0.573=2\)

\(\Leftrightarrow\left|x\right|=1.427\)

hay \(x\in\left\{1.427;-1.427\right\}\)

Vậy: \(x\in\left\{1.427;-1.427\right\}\)

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

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)

d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)

\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)

\(\Leftrightarrow x=\dfrac{1}{250}\)

Vậy: \(x=\dfrac{1}{250}\)

`a)sqrt{1-4x+4x^2}+5=x-2`

`<=>\sqrt{(2x-1)^2}=x-2-5`

`<=>|2x-1|=x-7(x>=7)`

`<=>[(2x-1=x-7),(2x-1=7-x):}`

`<=>[(x=-6(ktm)),(3x=8):}`

`<=>x=8/3(ktm)`

Vậy PTVN

`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`

`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`

`<=>4sqrt{x+3}=4`

`<=>sqrt{x+3}=1<=>x+3=1`

`<=>x=-2(tm)`

Vậy `S={-2}`

a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)

TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)

Vậy \(S=\varnothing\)

b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)

a. \(\sqrt{1-4x+4x^2}+5=x-2\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\)

\(\Leftrightarrow\left|1-2x\right|-x=-7\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2x-x=-7\\2x-1-x=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\x=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-6\end{matrix}\right.\)

b. ĐKXĐ: \(x\ge-3\)
\(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\)

\(\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}-6\sqrt{3+x}=4\)

\(\Leftrightarrow4\sqrt{3+x}=4\) \(\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\) ( thỏa mãn đk )

a) vì /x/>= 0 => x = { 2,5 ; -2,5 }

b) ko tìm đươc x thỏa mãn vì /x/ >= 0

c) /x/ = 2 + 0,573

<=> /x/ = 2,573

<=> x = { 2,573 ; -2,573 }

d) /x+ 1/3 / = -1+(-4 )

<=> /x+1/3 /= -5

vì /x+1/3 / luôn lớn hơn hoặc bằng 0 => ko tìm được x thỏa mãn

a. |x| = 2,5

=> \(x\in\left\{-2,5;2,5\right\}\).

b. |x| = -1,2

Mà |x| > 0 ( theo lí thuyết )

Vậy không tồn tại x thỏa mãn.

c. |x|+0,573=2

=> |x|=2-0,573

=> |x|=1,427

=> \(x\in\left\{-1,427;1,427\right\}\).

d. |x+1/3|-4=-1

=> |x+1/3|=-1+4

=> |x+1/3|=3

+) x+1/3=3

=> x=3-1/3

=> x=8/3

+) x+1/3=-3

=> x=-3-1/3

=> x=-10/3

Vậy \(x\in\left\{-\frac{10}{3};\frac{8}{3}\right\}\).