`#3107`

`a)`

`(x - 20)^3 = -64`

`\Rightarrow (x - 20)^3 = (-4)^3`

`\Rightarrow x - 20 = -4`

`\Rightarrow x = -4 + 20`

`\Rightarrow x = 16`

Vậy, `x = 16.`

`b)`

`9(x + 1) - 11(x + 7) = 105 - 1`

`\Rightarrow 9(x + 1) - 11(x + 7) = 104`

`\Rightarrow 9x + 9 - 11x - 77 = 104`

`\Rightarrow (9 - 11)x + (9 - 77) = 104`

`\Rightarrow -2x - 68 = 104`

`\Rightarrow -2x = 104 + 68`

`\Rightarrow -2x = 172`

`\Rightarrow x = 172 \div (-2)`

`\Rightarrow x = -86`

Vậy,` x = -86.`

a) Ta có ( x - 20 )³ = -64 = ( -4 )³

⇒ x - 20 = -4

⇒ x = 20 + ( -4 ) = 16

b) Ta có 9 ( x + 1 ) - 11 ( x + 7 ) = 105 - 1

⇒ 9x + 9 - 11x - 77 = 104

⇒ ( -2 ) . x - 68 = 104

⇒ ( -2 ) . x = 104 - 68 = 36

⇒ x = 36 : ( -2 ) = -18

a)

\(\left(x-20\right)^3=-64\)

\(\left(x-20\right)^3=-4^3\)

\(\Rightarrow x-20=-4\)

\(\Rightarrow x=-4+20=16\)

b)

\(9\left(x+1\right)-11\left(x+7\right)=105-1\)

\(9\left(x+1\right)-11\left(x+1+6\right)=104\)

\(9\left(x+1\right)-11\left(x+1\right)+66=104\)

\(\left(x+1\right).\left(9-11\right)=104-66\)

\(\left(x+1\right).\left(-2\right)=38\)

\(x+1=38:\left(-2\right)\)

\(x+1=-19\)

\(x=-19-1\)

\(x=-20\)

a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)

a, ĐKXĐ : \(x\ge1\)

Ta có ; \(PT\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.\sqrt{9}\sqrt{x-1}+24.\sqrt{\dfrac{1}{64}}\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\dfrac{1}{2}-\dfrac{3}{2}\sqrt{9}+24\sqrt{\dfrac{1}{64}}\right)=-17\)

\(\Leftrightarrow-\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x=290\left(TM\right)\)

Vậy ....

b, ĐKXĐ : \(x\ge3\)

Ta có : \(PT\Leftrightarrow x-3-7\sqrt{x-3}+12=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\) ( TM )

Vậy ..

a) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow-\sqrt{x-1}=-17\)

\(\Leftrightarrow x-1=17^2=289\)

hay x=290

Vậy: S={290}

b) Ta có: \(x-7\sqrt{x-3}+9=0\)

\(\Leftrightarrow x-7\sqrt{x-3}=-9\)

\(\Leftrightarrow x-3-2\cdot\sqrt{x-3}\cdot\dfrac{7}{2}+\dfrac{49}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow\left(\sqrt{x-3}-\dfrac{7}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=4\\\sqrt{x-3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\)

Vậy: S={19;12}

a \(ĐKXĐ:x\ge1\) 

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3^2}{2}\sqrt{x-1}+\dfrac{24}{8}\sqrt{x-1}=-17\Leftrightarrow\dfrac{1}{2}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\) \(\Leftrightarrow-4\sqrt{x-1}+3\sqrt{x-1}=-17\Leftrightarrow-\sqrt{x-1}=-17\Leftrightarrow\sqrt{x-1}=17\Rightarrow x-1=289\Leftrightarrow x=290\left(TM\right)\) b \(ĐKXĐ:x\ge3\) 

\(\Leftrightarrow x-3-7\sqrt{x-3}+12=0\Leftrightarrow\left(\sqrt{x-3}-3\right)\left(\sqrt{x-3}-4\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=3\\\sqrt{x-3}=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(TM\right)\\x=19\left(TM\right)\end{matrix}\right.\)

1)\(\left(2^5:2^3\right).2^x=64\)

\(\Rightarrow2^{5-3+x}=2^6\)

\(\Rightarrow2^{2+x}=2^6\)

\(\Rightarrow.2^22^x=2^6\)

\(\Rightarrow2^x=2^6:2^2\)

\(\Rightarrow2^x=2^4\Rightarrow x=4\)

2)Tính:

\(F=3^0+3^1+...+3^9\)

\(\Rightarrow3F=3\left(3^0+3^1+...+3^9\right)=3+3^2+3^3+...+3^{10}\)

\(3F-F=3+3^2+...+3^{10}-3^0-3^1-...-3^9\)

\(2F=3^{10}-3^0=3^{10}-1\)

\(F=\frac{3^{10}-1}{2}\)

2 

ta có : F = 1 + 3 + 3² + ..... + 3⁹

=> 3F = 3 + 3² + 3³ +..... + 3¹⁰ 

=> 3F - F = 3¹⁰ - 1

=> 2F = 3¹⁰ - 1

=> F = \(\frac{3^{10}-1}{2}\)

(2^5:2^3)*2^x=64

2^2. 2^x = 2^6

2^x= 2^6 : 2^2

2^x = 2^4

=> x= 4

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

hơi khó anh mai ơi !

hơi bị khó... chờ mình ghi lại để hỏi cô!!!

dấu này"/" là chia hay phần

A = ( 4/4 + 2/3 ) - ( 51/3 - 6/5 ) - ( 6 - 7/4 + 3/2 )

Sau đó quy đồng rồi trừ cả là đc 

B tương tự 

C=13/15 

D cx thế . Bạn tự vận dụng đi . Xl vì ko giải đc . Mik đang gấp

fgv  vttf bv  vb v bv g

2. 2(x-1) +3 2-x) =- 1

\(\Leftrightarrow2x-2+6x-3=-1\)

\(\Leftrightarrow8x-5=-1\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2

3. ( n² + 3 ) chia hết cho ( n - 1)

\(\Leftrightarrow n^2-1+4⋮n-1\Leftrightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)

Vì n thuộc Z => ( n-1) ( n+1) thuộc Z

\(\Rightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\Leftrightarrow4⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Phần còn lại bn tự làm