a, -19 - x = -20

           x = -19 - (-20)

           x = -19 + 20

           x = 1

b, 5x - 6 = 3x + 12

    5x - 6 - 3x = 12

    5x - 3x      = 12 + 6

    (5 - 3)x      = 18

    2x             = 18

    x              = 18 : 2

    x              = 9

c, 15 - 3 (x - 1) = 8 - 2x

    15 - 3 (x - 1) + 2x = 8

          -3x - 3 - 2x = 8 - 15

          -3x - 3 - 2x = -7

          -3x - 2x - 3 = 7

          -3x - 2x       = 7 + 3

           (-3 - 2) x      = 10

                   -5x     = 10

                      x      = 10 : (-5)

                      x      = -2

d, (5x - 6)² = 16

    (5x - 6)² = 4²

=> 5x - 6 = 4

     5x      = 4 + 6

     5x      = 10

       x      = 10 : 5

       x     = 2

f, 26 - | x + 9 | = 13

          | x + 9 | = 26 - 13

=>      | x + 9 | = 13

=>        x + 9 = +- 13

* Với x + 9 = 13

         x      = 13 - 9

         x      = 4

* Với x + 9 = -13

         x      = -13 - 9

         x      = -22

Vậy x = {4;-22}

e, | 3 + x | = 19

=>  3 + x = +- 19

* Với 3 + x = 19

              x = 19 - 3

              x = 16

* Với 3 + x = -19

              x = -19 - 3

              x = -22

Vậy x = {16;-22}

a, X = -19+20=1

b, (5-3)X = 18

       2X   = 18

=>    X    = 9

c, 3X + 3 -2X = 7

         X+3     =7

           X       = 4

f, |X+9| = 13

ta có 2 trường hợp:

TH1: X+9 = 13

=> X= 4

TH2 : X+9 = -13 

=> X= -22

e, ta có 2 trường hợp:

TH1: 3+X = 19

=> X= 16

TH2: 3+X = -19

=> X= -22

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

1. -x+20 = -(-15)-8+13

=> -x=15-8+13-20

=> -x=0

=> x=0

2. -(-10)+x=-13+(-9)+(-6)

=> 10+x=-13-9-6

=> x = -13-9-6-10

=> x = -38

3. 8-(-12)+10=-(-14)-x

=> 8+12+10=14-x

=> x = 14-8-12-10

=> x = -16

4. -(+12)+(-x)-(-3)=5-(-7)

=> -12-x+3=5+7

=> -x=5+7+12-3

=> -x=21

=> x=-21

5. 14-x+(-10)=-(-9)+(+15)

=> 14-x-10=9+15

=> -x=9+15-14+10

=> -x=20

=> x=-20

6. 12-(-17)+(-3)=-5+x

=> 12+17-3+5=x

=> x=31

7. x-(-19)-(+32)=14-(+16)

=> x+19-32=14-16

=> x=14-16+32-19

=> x=11

8. x-|-15|-|7|=-(-9)+|-5|

=> x-15-7=9+5

=> x=9+5+7+15

=> x=36

9. 15-x+17=13-(-21)

=> 15-x+17=13+21

=> -x=13+21-15-17

=> -x=2

=> x=-2

10. -|-5|-(-x)+4=3-(-25)

=> -5+x+4=3+25

=> x=3+25-4+5

=> x=29

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

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

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

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

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

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

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

Tìm x . biết : 

\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\)

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)

\(1-x=\frac{2}{5}:\frac{2}{5}=1\)

\(x=1-1=0\)

\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)

\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)

\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)

học tốt nha

a)\(\frac{x}{5}=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{2\times5}{5}=2\)

Vậy .............

b) \(\frac{3}{8}=\frac{6}{x}\)

\(\Leftrightarrow x=\frac{8\times6}{3}=16\)

Vậy ................

c) \(\frac{1}{9}=\frac{x}{27}\)

\(\Leftrightarrow x=\frac{1\times27}{9}=3\)

Vậy ................

d) \(\frac{4}{x}=\frac{8}{6}\)

\(\Leftrightarrow x=\frac{4\times6}{8}=3\)

Vậy ............

e) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)

\(\Leftrightarrow3x+6=-4x+20\)

\(\Leftrightarrow3x+4x=20-6\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

Vậy .............

f) \(\frac{x}{-2}=\frac{-8}{x}\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x=\pm4\)

Vậy ...............

\(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\times5}{5}=2\)

\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{1\times27}{9}=3\)

\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{8\times6}{8}=6\)

\(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow3x+4x=20-6\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=2\)

câu đầu bn suy ra x = 2 luôn nha, viết lại đề suy ra luôn