Bài 1 :

a) 7^2x-1 = 343

=> 7^2x-1 = 7³

=> 2x - 1 = 3 => 2x = 4 => x = 2

b) (7x - 11)³ = 2⁵.3² + 200

=> (7x - 11)³ = 32.9 + 200

=> (7x - 11)³ = 488

xem kĩ lại đề này :vvv

c) 174 - (2x - 1)² = 5³

=> (2x - 1)² = 174 - 5³

=> (2x - 1)² = 174 - 125 = 49

=> (2x - 1)² = (\(\pm\)7)²

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

Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)

Bài 2 :

a) x⁵ = 32 => x⁵ = 2⁵ => x = 2

b) (x + 2)³ = 27

=> (x + 2)³ = 3³

=> x + 2 = 3 => x = 3 - 2 = 1

c) (x - 1)⁴ = 16

=> (x - 1)⁴ = 2⁴

=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)

d) (x - 1)⁸ = (x - 1)⁶

=> (x - 1)⁸ - (x - 1)⁶ = 0

=> (x - 1)⁶ [(x - 1)² - 1] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)

+) x - 1 = 1 => x = 2 ( tm)

+) x - 1 = -1 => x = 0 ( tm)

Vậy x = 1,x = 2,x = 0

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

Bài 1:

<=>7[3(-x)]-12(x-5)=-3(11x-20)

=>-3(11x-20)=5

=>-33x=-55

=>-11.3x=-11.5 (rút gọn -11)

=>3x=5

\(\Rightarrow x=\frac{5}{3}\)

Đã duyệt

bài 1:

<=>7[3(-x)]-12(x-5)=-3(11x-20)

=>-3(11x-20)=5

=>-33x=-55

=>-11.3x=-11.5 (rút gọn -11)

=>3x=5

=>x=\(\frac{5}{3}\)

bài 1:

<=>7[3(-x)]-12(x-5)=-3(11x-20)

=>-3(11x-20)=5

=>-33x=-55

=>-11.3x=-11.5 (rút gọn -11)

=>3x=5

=>x=\(\frac{5}{3}\)

đậu phộng thằng Phan Bá Lộc chép bài ông

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

(x+2).(y-3)=-3=-1.3=1.(-3)

Vì x,y thuộc Z nên ( x+2) và (y+3) thuộc Z

Ta có bảng:

Vậy nếu x = - 3 thì y = 0

       nếu x = -1 thì y =- 6

       nếu x = - 5 thì y = - 2

       nếu x = 1 thì y = - 4

sao ko làm hết  bài

a) 1 - 2 + 3 - 4 + 5 - 6 + ... + 159 - 160

= (1 - 2) + (3 - 4) + (5 - 6) + ... + (159 - 160)

= (-1) + (-1) + (-1) + ... + (-1)

       có 80 số -1

= (-1) . 80

= -80

1 - 2 + 3 - 4 + 5 - 6 + ......... + 159 - 160 ( có 160 số )

= - 1 + ( - 1 ) + ( - 1 ) + .......... + ( - 1 )    ( có 80 số - 1 )

= - 1 . 80

= - 80 

\(12-\left|x+7\right|=12\)

\(\Rightarrow\left|x+7\right|=12-12\)

\(\Rightarrow\left|x+7\right|=0\)

\(\Rightarrow x+7=0\)

\(\Rightarrow x=0-7\)

\(\Rightarrow x=-7\)

bài 1: x.(x+7) = 0

Th1:x=0              Th2:x+7=0

                          =>x=-7

bài 2 (x+12).(x-3)= 0

Th1:x+12=0                                         Th2:x-3=0

=>x=-12                                                =>x=3

bài 3 (-x+5).(3-x)=0

Th1 (-x)+5=0                                          Th2:3-x=0

=>-x=-5                                                  =>x=3

bài 4 x.(2+x).(7-x)=0

Th1:x=0                                               Th3:7-x=0

Th2:2+x=0                                             =>x=7

=>x=-2

bài 5 (x-1).(x+2).(-x-3)=0

Th1:x-1=0                                               Th2:x+2=0

=>x=1                                                   =>x=-2

Th3:-x-3=0

=>-x=-3