Có nhiều 3 cách viết em nhé!
C = { trâu; bò, chó}

C = { trâu; bò, mèo}

C = { trâu; bò, gà}

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

\(\Rightarrow x^2-4x+4-x^2+3x-9=0\)

\(\Rightarrow-x-5=0\)

=> x = -5

b) \(\left(5x-2\right)^2=\left(4-x\right)^2\)

\(\Rightarrow25x^2-10x+4-16+8x-x^2=0\)

\(\Rightarrow24x^2-2x-12=0\)

\(\Rightarrow12x^2-x-6=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)

c) \(x^2-4x-5=0\)

=> (x - 5).(x + 1) = 0

=> x = 5 hoặc x = -1

a) \(=\left(a+b\right)^2=a^2+2ab+b^2\)

b) \(=\left(a-b\right)^2=a^2-2ab+b^2\)

d) \(=\left(a+b\right)\left(a+b\right)^2=\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

e) \(=\left(a-b\right)\left(a-b\right)^2=\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)

g) \(\left(a-b\right)\left(a^2+ab+b^2\right)=a^3-b^3\)

Với x < 0; |x| là 1 số dương => |x| là số đối của x

=> |x| = -x

bạn viết câu hỏi dưới dạng trực quan để mn dễ hiểu nhé!

Baif mấy em ơi

ĐK: x ≥ 0

\(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{1}{x-\sqrt{x}+1}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)

b) \(N=\dfrac{8}{9}\)

\(\Rightarrow\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}=\dfrac{8}{9}\)

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

\(\Rightarrow3\sqrt{x}-2x+2\sqrt{x}-2=0\)

\(\Rightarrow5\sqrt{x}-2x-2=0\)

\(\Rightarrow-2x+4\sqrt{x}+\sqrt{x}-2=0\)

\(\Rightarrow-2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\left(\sqrt{x}-2\right)\left(-2\sqrt{x}+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{4}\end{matrix}\right.\left(t.m\right)\)