Bài 2:

a: =>x^2=60

=>\(x=\pm2\sqrt{15}\)

b: =>2^2x+3=2^3x

=>3x=2x+3

=>x=3

c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)

\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)

=>1/2x-2=4

=>1/2x=6

=>x=12

a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)

\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)

=>x=2 hoặc x=-3/2

b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)

=>2x+5=1/44 hoặc 2x+1=-1/44

=>x=-219/88 hoặc x=-221/88

1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna

Đăng ít thôi

1. B = (-3).|-7| - (-4).|5| + \(\dfrac{1}{3}.\left|-9\right|\)
= -21 + 20 + 3
= 2
@Shine Anna

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)

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

                \(\dfrac{3}{2}+x=\dfrac{3}{2}\)

                       \(x=\dfrac{3}{2}-\dfrac{3}{2}\)

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)

<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)

<=>\(4x-17=0\)

<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)

b) =>\(\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\3-\dfrac{1}{2}x=0\\1+\dfrac{3}{4}:x=0\end{matrix}\right.=>\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}x=3\\\dfrac{3}{4x}=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\x=6\\x=\dfrac{-3}{4}\end{matrix}\right.\)

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26

Bài 1:

\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)

Bài 2.

Sửa đề

a) \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)

Giải

Ta sẽ phân tích vế phải

VP = \(\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)

VP = \(\dfrac{a\left(x^2-4\right)+bx\left(x+2\right)+cx\left(x-2\right)}{x\left(x^2-4\right)}\)

VP = \(\dfrac{ax^2-4a+bx^2+2bx+cx^2-2cx}{x\left(x^2-4\right)}\)

VP = \(\dfrac{x^2\left(a+b+c\right)+2x\left(b-c\right)-4a}{x\left(x^2-4\right)}\)

Tương tự , ta cũng sẽ phân tích VT

VT = \(\dfrac{2x.5-4}{x\left(x^2-4\right)}\)

Đồng nhất hai VT và VP , ta có :

\(x^2\left(a+b+c\right)+2x\left(b-c\right)-4a=2.5x-4\)

* a + b + c = 0 => 1 + c + 5 + c = 0 => 2c = - 6 => c = - 3

* b - c = 5 => b = c + 5 => b = - 3 + 5 => b = 2

* a = 1

Vậy , a = 1 ; b = 2 ; c = -3

b) Ta sẽ phân tích VP

VP = \(\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)

VP = \(\dfrac{a\left(x^2+x+1\right)+\left(bx+c\right)\left(x-1\right)}{x^3-1}\)

VP = \(\dfrac{ax^2+ax+a+bx^2-bx+cx-c}{x^3-1}\)

VP = \(\dfrac{x^2\left(a+b\right)+x\left(a-b+c\right)+a-c}{x^3-1}\)

Đồng nhất VP và VT , ta được :

\(x^2\left(a+b\right)+x\left(a-b+c\right)+a-c=1\)

* a + b = 0 => a = - b => b = \(-\dfrac{1}{3}\)

* a - b + c = 0 => a + a + a - 1 = 0 => 3a = 1 => a = \(\dfrac{1}{3}\)

* a - c = 1 => c = a - 1 => c = \(\dfrac{1}{3}\) - 1 = \(-\dfrac{2}{3}\)

Vậy , a = \(\dfrac{1}{3}\) ; b = \(-\dfrac{1}{3}\); c = \(-\dfrac{2}{3}\)

Bài 1 bạn Giang làm rồi thì thôi nhé

Kiểm tra giùm mk câu a bài 2 nha!!! ĐỀ BÀI!!!