a) \(\left(\dfrac{1}{2}m+3\right)^3=\dfrac{m^3}{8}+\dfrac{9m^2}{4}+\dfrac{27m}{2}+9\)

b) \(2\left(m+\dfrac{1}{2}\right)=2m+1\)

c) \(\left(2\sqrt{x}+1\right)^2=4x+4\sqrt{x}+1\)

d) \(\left(2\sqrt{2}+\dfrac{1}{4}\right)^2=8+\sqrt{2}+\dfrac{1}{16}=\dfrac{129}{16}+\sqrt{2}\)

chi tiết câu c

c) \(\left(2\sqrt{x}+1\right)^2=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1^2=4x+4\sqrt{x}+1\)

a) \(\left(\dfrac{1}{2}m+3\right)^3\)

\(=\left(\dfrac{1}{2}m\right)^3+3\cdot\left(\dfrac{1}{2}m\right)^2\cdot3+3\cdot\dfrac{1}{2}m\cdot3^2+3^3\)

\(=\dfrac{1}{8}m^3+\dfrac{9}{4}m^2+\dfrac{27}{2}m+27\)

c) Ta có: \(\left(2\sqrt{x}+1\right)^2\)

\(=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1^2\)

\(=4x+4\sqrt{x}+1\)

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

23/12

1/3

a.x=23/12

b.x=1/3

x = 5/3+1/4

x = 23/12

x = 1/2 x 2/3

x= 1/3

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

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

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

Câu 2: 

a: \(\Leftrightarrow x+2\in\left\{3;9\right\}\)

hay \(x\in\left\{1;7\right\}\)

Thi à :)?

Hai bài bị trùng nhau nên các bạn nhìn ảnh hay văn bản đều như nhau ạ

c: =>x+2>0

hay x>-2

d: =>-4<=x<=3

e: =>\(x\in\varnothing\)

f: \(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -6\end{matrix}\right.\)

1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)

2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)

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

\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)

3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)

\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

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

A. (Theo mình là -4xy thì mới rút gọn được)

B = (6x + y)^2

C = (2x)^2 = 4x^2

1. (x-3)²

2. (3y+2x)²

3. (1/5x-8y)(1/5x+8y)

4. (x-2y)(x²+2xy+4y²)

5. (4x-3-x-1)(4x-3+x+1)

(3x-4)(5x-2)

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

n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)

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

Mik cần gấp