Câu 6: Khôg có cau nào đúng

Câu 7: C

Câu 8: B

Câu 9: B

Câu 10: D

6Không có đáp án nào đúng x=11/4

7C

8B

9B

10D

a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)

\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)

=>2x=44/15

hay x=22/15

a. x × 5 = 20

x = 20 : 5

x = 4

b. x : 4 = 8 – 3

x : 4 = 5

x = 5 × 4

x= 20

a) x+3/22=-27/121.11/9

⇒ x+3/22=-3/11

⇒ x=-9/22

b) (3/2/20-2/1/20)x+2/7=3/5

⇒ (31/10-41/20)x=11/35

⇒ 21/20x=11/35

⇒x=44/147

c) |3/2x-5|-1/2=-1/4

⇒ |3/2x-5|=1/4

⇒ hoặc 3/2x-5=1/4⇒3/2x=21/4⇒x=7/2

hoặc 3/2x-5=-1/4⇒3/2x=19/4⇒x=19/6

-2.(x-5)-18=20-3.(x+4)

<=> -2x+10-18=20-3x-12

<=>-2x-8=8-3x

<=>-2x+3x=8+8

<=>x=16

Vậy x=16

Học tốt!!!

#Bo_lienminhmeovatrau

tìm x biết

-2.(x-5)-18=20-3.(x+4)

-2x+10-18=20-3x-12

-2x+3x=20-12-10+18

        x=16

Vậy x=16

-2(x-5)-18=20-3(x+4)

<=> -2x+10-18=20-3x-12

<=> -2x-8=-3x+8

<=> -2x-8+3x-8=0

<=> x-16=0

<=> x=16

\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)

\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)

\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)

\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)

Ta có:

3/4<|3/20-x|<4/5

=>Ta có 2 trường hợp sau:

TH1:

3/4<3/20-x<4/5

=>3/4+3/20<3/20-x+3/20<4/5+3/20

=>9/10<x<19/20

TH2:

-3/4>-(3/20-x)>--4/5

=>-3/4+3/20>-3/20+x+3/20>-4/5+3/20

=>-3/5>x>-13/20

Vậy 9/10<x<19/20 hoặc -3/5>x>-13/20

Ta có: \(20\%x-x+\dfrac{1}{5}=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-4}{5}x=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{15}{20}-\dfrac{4}{20}=\dfrac{11}{20}\)

\(\Leftrightarrow x=\dfrac{11}{20}:\dfrac{-4}{5}=\dfrac{11}{20}\cdot\dfrac{5}{-4}=\dfrac{-55}{80}=\dfrac{-11}{16}\)

Vậy: \(x=-\dfrac{11}{16}\)