1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

Bài 1: 

a: x/-2=-18/x

=>x²=36

=>x=6 hoặc x=-6

b: x/2+x/5=17/10

=>7/10x=17/10

hay x=17/7

y+y x 1/3 : 2/9 + y :2/7 = 252

y x 1 + y x (1/3 x 9/2 ) + y x 7/2 = 252

y x 1/1 + y x 3/2 + y x 7/2 = 252

y x (1/1 + 3/2 + 7/2 ) = 252

y x 6 = 252

y = 252 : 6

y = 42

Bài 1: 

Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)

Vậy: \(x=\dfrac{4}{65}\)

Bài 2: 

a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)

\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)

\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)

\(=\dfrac{5366}{527}\)

a)\(\frac{5}{9}\times x-1=\frac{8}{9}\)

\(\frac{5}{9}\times x=\frac{8}{9}+1\)

\(\frac{5}{9}\times x=\frac{17}{9}\)

\(x=\frac{17}{9}:\frac{5}{9}\)

\(x=\frac{17}{5}\)

b)\(x+x\times\frac{1}{3}:\frac{2}{9}+x:\frac{2}{7}=252\)

\(x+x\times\frac{3}{2}+x:\frac{2}{7}=252\)

\(x.1+x.\frac{3}{2}+x.\frac{7}{2}=252\)

\(x\left(1+\frac{3}{2}+\frac{7}{2}\right)=252\)

\(x\times6=252\)

\(x=252:6\)

\(x=42\)

1: Thay \(x=\dfrac{1}{25}\) vào C, ta được:

\(C=\left(\dfrac{1}{5}+2\right):\left(\dfrac{1}{5}-3\right)=\dfrac{11}{5}:\dfrac{-14}{5}=-\dfrac{11}{14}\)

2: Để C=-2 thì \(\sqrt{x}+2=-2\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow\sqrt{x}+2+2\sqrt{x}-6=0\)

\(\Leftrightarrow3\sqrt{x}=4\)

hay \(x=\dfrac{16}{9}\)

Để \(C=\dfrac{7}{5}\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{7}{5}\)

\(\Leftrightarrow7\sqrt{x}-21=2\sqrt{x}+10\)

\(\Leftrightarrow5\sqrt{x}=31\)

hay \(x=\dfrac{961}{25}\)

giải chi tiết giùm mình nhé 

nếu mà bạn giải xong rồi thì bạn nói cho mình biết nhen 

Ta có : (x + 2) + (x + 5) + (x + 8) + ...... + (x + 32) = 198

<=> (x + x + x + ...... + x) + (2 + 5 + 8 + ....... + 32) = 198

<=> 11x + 187 = 198

=> 11x = 198 - 187

=> 11x = 11

=> x = 1

a . 

 \(x+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)

\(x+x\times\frac{3}{2}+x=252\times\frac{2}{7}\)

\(x+x\times\frac{3}{2}+x=72\)

\(x=72\div\frac{3}{2}-3x\)

\(x=48-3x\)

\(x=45\)