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b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
\(B=\sqrt{16a^4}+6a^2=4a^2+6a^2=10a^2\)\(A=\sqrt{49a^2}+3a=7a+3a=10a\)
\(C=4x-\sqrt{\left(x^2-4x+4\right)}=4x-\sqrt{\left(x-2\right)^2}=4x-x+2=3x+2\)
\(E=\sqrt{y^2+6y+9}-\sqrt{y^2-6y+9}=\sqrt{\left(y+3\right)^2}-\sqrt{\left(y-3\right)^2}=\left|y+3\right|-\left|y-3\right|=y+3-y+3=6\)
\(D=\dfrac{a-b}{\sqrt{a}-\sqrt{b}}=\dfrac{\left(a-b\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{a\sqrt{a}+a\sqrt{b}-b\sqrt{a}-b\sqrt{b}}{a-b}=\dfrac{\sqrt{a}\cdot\left(a-b\right)+\sqrt{b}\cdot\left(a-b\right)}{a-b}=\dfrac{\left(a-b\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\sqrt{a}+\sqrt{b}\)
Bài 1:
a) Ta có: \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}\right)^2-1^2\)
\(=x-1\)
b) Ta có: \(\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}\right)^3+1^3\)
\(=x\sqrt{x}+1\)
c) Ta có: \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(=2x-2\sqrt{x}+\sqrt{x}-1\)
\(=2x-\sqrt{x}-1\)
Bài 2: Tìm x
a) Ta có: \(\sqrt{9x^2+6x+1}=3x-2\)
\(\Leftrightarrow\left|3x+1\right|=3x-2\)(*)
Trường hợp 1: \(x\ge\frac{-1}{3}\)
(*)\(\Leftrightarrow3x+1=3x-2\)
\(\Leftrightarrow3x+1-3x+2=0\)
\(\Leftrightarrow3=0\)(vô lý)
Trường hợp 2: \(x< \frac{-1}{3}\)
(*)\(\Leftrightarrow-3x-1=3x-2\)
\(\Leftrightarrow-3x-1-3x+2=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\frac{1}{6}\)(loại)
Vậy: \(S=\varnothing\)
b)Trường hợp 1: \(x\ge0\)
Ta có: \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4(nhận)
Vậy: S={x|x>4}
Minh bi nham dau bai, chi co 1 thua so \(\dfrac{2}{x}\) thoi nhe!