`#040911`

`a,`

`15 + 25 \div (2x - 1) = 20`

`\Rightarrow 25 \div (2x - 1) = 20 - 15`

`\Rightarrow 25 \div (2x - 1) = 5`

`\Rightarrow 2x - 1 = 25 \div 5`

`\Rightarrow 2x - 1 = 5`

`\Rightarrow 2x = 6`

`\Rightarrow x = 3`

Vây, `x = 3.`

`b,`

\(3^{x-1}+2\cdot3^x=21\)

`\Rightarrow 3^x \div 3 + 2. 3^x = 21`

`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`

`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`

`\Rightarrow 3^x . \frac{7}{3} = 21`

`\Rightarrow 3^x = 21 \div \frac{7}{3}`

`\Rightarrow 3^x = 9`

`\Rightarrow 3^x = 3^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

`c,`

\(2^{x-3}+2^{x+1}=17\)

`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`

`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`

`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`

`\Rightarrow 2^x . \frac{17}{8} = 17`

`\Rightarrow 2^x = 17 \div \frac{17}{8}`

`\Rightarrow 2^x = 8`

`\Rightarrow 2^x = 2^3`

`\Rightarrow x = 3`

Vậy, `x = 3`

`d,`

\(5^x-5^{x-1}=20\)

`\Rightarrow 5^x - 5^x \div 5 = 20`

`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`

`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`

`\Rightarrow 5^x . \frac{4}{5} = 20`

`\Rightarrow 5^x = 20 \div \frac{4}{5}`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

\(a.25:\left(2x-1\right)=5\)

\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)

\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)

\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)

\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)

\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)

\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)

\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)

\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

`a,4x-10=0   `

`<=> 4x=10`

`<=>x=10/4`

`<=>x=5/2`

`b, 7-3x=9-x     `

`<=>-3x+x=9-7`

`<=>-2x=2`

`<=>x=-1`

`c, 2x-(3-5x) = 4(x+3)`

`<=>2x-3+5x=4x+12`

`<=>2x+5x-4x=12+3`

`<=>3x=15`

`<=>x=5`

`d, 5-(6-x)=4(3-2x)     `

`<=>5-6+x=12-8x`

`<=>x+8x=12-5+6`

`<=>9x=13`

`<=>x=13/9`

`e, 4(x+3)=-7x+17   `

`<=>4x+12=-7x+17`

`<=>4x+7x=17-12`

`<=>11x=5`

`<=>x=5/11`   

`f, 5(x-3) - 4=2(x-1)+7`

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`g, 5(x-3)-4=2(x-1)+7       `

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`h,4(3x-2)-3(x-4)=7x+20`

`<=>12x-8-3x+12=7x+20`

`<=>12x-3x-7x=20+8+12`

`<=>2x=40`

`<=>x=20`

1. 3y = 0

=> y = 0

2. 1+x = 0

<+ x = -1

3.

\(1-2t=0\)

\(\Leftrightarrow2t=1\)

\(\Leftrightarrow\dfrac{1}{2}\)

4. 2x +x + 3 =0

\(\Leftrightarrow3x+3=0\)

\(\Leftrightarrow x=-3\)

5.

\(25x-20=0\)

\(\Leftrightarrow25x=20\)

\(\Leftrightarrow x=\dfrac{4}{5}\)

7.

2x-3 = x+5

<=> 2x - x = 5+3

<=> x = 8

8.

x-8=2x+3

<=> x - 2x = 3+8

<=> -x = 11

<=> x = -11

9. 17-2x = 3x-5

<=> -2x-3x = -5-17

<=> -5x = -22

<=> x = \(\dfrac{22}{5}\)

10.

2x+x+22=0

<=> 3x+22=0

<=> 3x = -22

<=> x = \(\dfrac{-22}{3}\)

Mấy bài kia tự giải tương tự nhá!!!

11.

x-3=3x+5

<=> x-3x=5+3

<=> -2x=8

<=> x = -4

12. 15x + 1/2 = 0

<=> 15x = -1/2

<=> x = -1/30

13.

3x+1 = 7x-11

<=> 3x-7x = -11-1

<=> -4x=-12

<=> x = 3

14.

5-3x = 6x + 7

<=> -3x-6x = 7-12

<=> -9x = -5

<=> x = 5/9

15.

11-2x=x-1

<=> -2x-x = -1-11

<=> -3x = -12

<=> x = 4

16.

15 - 8x = 9-5x

<=> -8x+5x = 9 - 15

<=> -3x = -6

<=> x=2

17.

25-3x=4x-5

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

<=> -7x = -30

<=> x = 30/7

18.

0,25x+1,5=0

<=> 0,25x = -1,5

<=> x = -6

19.

3x-5/6 = 1/2

<=> 3x - 5/6 = 3/6

<=> 3x-5=3

<=> 3x = 8

<=> x = 8/3

20.

-2x + 14 =0

<=> -2x = -14

<=> x = 7

a)x=-17

b)x=9/10

c)x=4\(\frac{1}{3}\)

tick đi giải chi tiết cho

a)Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau

7x+35/3=2x+6/1=>(7x+35)1=3(2x+6)

=>x=-17

b)Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau

17x+19/20=27x+10/20=>(17x+19)20=20(27x+10)

c)<=>(x-2)^3+(x-4)^3+(x-7)^3+(-3)(x-2)(x-4)(x-7)=19(3x-13)

=>19(3x-13)=0

rút gọn 57x=247

=>19.3x=19.13

=>3x=13

=>x=13/3

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

b)  

         \(\frac{x+1}{4}-\frac{2x-1}{5}+\frac{2x+1}{2}=\frac{27x+10}{20}\) 

<=>   \(\frac{5\left(x+1\right)}{20}-\frac{4\left(2x-1\right)}{20}+\frac{10\left(2x+1\right)}{20}=\frac{27x+10}{20}\)

<=>  \(5\left(x+1\right)-4\left(2x-1\right)+10\left(2x+1\right)=27x+10\)

<=>   \(5x+5-8x+4+20x+10=27x+10\)(Bước này có thể bỏ)

<=>   \(10x=-9\)

<=>  \(x=-\frac{9}{10}\)

Vậy tập nghiệm của phương trình: S=\(\left(-\frac{9}{10}\right)\) ( thay ngoặc tròn thành ngoặc nhọn )

P/s: Tớ chỉ biết làm như thế thôi!! :))) 

Áp dụng : (A + B)³ = A³ + 3A²B + 3AB² + B³

11) \(\left(x^2+\frac{3}{xy}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{3}{xy}+3\cdot x^2\cdot\left(\frac{3}{xy}\right)^2+\left(\frac{3}{xy}\right)^3\)

\(=x^6+3\cdot x^4\cdot\frac{3}{xy}+3\cdot x^2\cdot\frac{9}{x^2y^2}+\frac{27}{x^3y^3}\)

\(=x^6+\frac{9x^4}{xy}+\frac{27\cdot x^2}{x^2y^2}+\frac{27}{x^3y^3}\)

\(=x^6+\frac{9x^3}{y}+\frac{27}{y^2}+\frac{27}{x^3y^3}\)

12) \(\left(x^2+\frac{2}{x}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{2}{x}+3\cdot x^2\cdot\left(\frac{2}{x}\right)^2+\left(\frac{2}{x}\right)^3\)

\(=x^6+3\cdot x^4\cdot\frac{2}{x}+3\cdot x^2\cdot\frac{4}{x^2}+\frac{8}{x^3}\)

\(=x^6+\frac{6\cdot x^4}{x}+\frac{12\cdot x^2}{x^2}+\frac{8}{x^3}\)

\(=x^6+6x^3+12+8x^3\)

13) \(\left(3y+\frac{x}{2}\right)^3=\left(3y\right)^3+3\cdot3y^2\cdot\frac{x}{2}+3\cdot3y+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3\)

\(=27y^3+\frac{9y^2\cdot x}{2}+9y+\frac{x^2}{4}+\frac{x^3}{8}\)

14) \(\left(1\frac{1}{2}xy+1\right)^3=\left(\frac{3}{2}xy+1\right)^3=\left(\frac{3}{2}xy\right)^3+3\cdot\left(\frac{3}{2}xy\right)^2\cdot1+3\cdot\frac{3}{2}xy\cdot1^2+1^3\)

\(=\frac{27}{8}x^3y^3+3\cdot\frac{9}{4}x^2y^2+\frac{9}{2}xy+1\)

\(=\frac{27}{8}x^3y^3+\frac{27}{4}x^2y^2+\frac{9}{2}xy+1\)

15) \(\left(\frac{x^2}{2}+\frac{2}{y}\right)^3=\left(\frac{x^2}{2}\right)^3+3\cdot\left(\frac{x^2}{2}\right)^2\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\left(\frac{2}{y}\right)^2+\left(\frac{2}{y}\right)^3\)

\(=\frac{x^6}{8}+3\cdot\frac{x^4}{4}\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\frac{4}{y^2}+\frac{8}{y^3}\)

\(=\frac{x^6}{8}+\frac{3x^4}{2y}+\frac{6x^2}{y^2}+\frac{8}{y^3}\)

Còn 5 bài cuối áp dụng tương tự như thế :)