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Ta có:\(A=\left(\dfrac{a+4\sqrt{a}+4}{a+2\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right):\left(\dfrac{\sqrt{a}-4}{a-2\sqrt{a}}-\dfrac{3\sqrt{a}+6}{4-a}\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right]:\left[\dfrac{\sqrt{a}-4}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{3\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right]\)
\(=\dfrac{a-4-a-2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}:\dfrac{\sqrt{a}-4+3\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\dfrac{-4-2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}.\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{4\sqrt{a}-4}=\dfrac{-2-\sqrt{a}}{2\sqrt{a}-2}\)
Ta có: \(A=\left(\dfrac{a+4\sqrt{a}+4}{a+2\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right):\left(\dfrac{\sqrt{a}-4}{a-2\sqrt{a}}-\dfrac{3\sqrt{a}+6}{4-a}\right)\)
\(=\left(\dfrac{\sqrt{a}+2}{\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right):\left(\dfrac{\sqrt{a}-4}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{3}{\sqrt{a}-2}\right)\)
\(=\dfrac{a-4-a}{\sqrt{a}\left(\sqrt{a}-2\right)}:\dfrac{\sqrt{a}-4+3\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\dfrac{-4}{4\left(\sqrt{a}+1\right)}=\dfrac{-1}{\sqrt{a}+1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)
\(P=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}-\dfrac{4}{\sqrt{ }a}\right)\)
\(=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
\(P=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\) (ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\))
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2-\sqrt{a}-2\right)\left(\sqrt{a}-2+\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\right)\)
\(=\dfrac{-4.2\sqrt{a}}{\sqrt{a}}\)
\(=-8\)
#YM
Lời giải:
ĐKXĐ: $a\geq 0; a\neq 4$
\(A=\left[\frac{\sqrt{a}(\sqrt{a}-2)-\sqrt{a}(\sqrt{a}+2)}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}\right].(\sqrt{a}+2)\)
\(=\frac{-4\sqrt{a}+4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{1}{2-\sqrt{a}}\)
\(A=2\cdot\sqrt{9+4\sqrt{5}}+\sqrt{5}-3\sqrt{5}\)
=2(căn 5+2)-2căn 5
=4
Đặt \(\sqrt[3]{a}=x;\sqrt[3]{b}=y\)
=>\(Q=\dfrac{x^4+x^2y^2+y^4}{x^2+xy+y^2}\)
\(=\dfrac{x^4+2x^2y^2+y^4-x^2y^2}{x^2+xy+y^2}\)
\(=\dfrac{\left(x^2+y^2\right)^2-\left(xy\right)^2}{x^2+xy+y^2}=\dfrac{\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)}{x^2+xy+y^2}\)
\(=x^2-xy+y^2\)
\(=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}\)
\(P=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)
\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\frac{4\sqrt{a}-4}{a-4}\)
\(=\frac{a+5\sqrt{a}+6-\left(a-3\sqrt{a}+2\right)-\left(4\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)
Bài 4:
Để hai đường song song thì 2m-1=5
=>2m=6
=>m=3
Bài 3:
a: 4x-3y=2 và 4x+3y=-18
=>8x=-16 và 4x-3y=2
=>x=-2 và 3y=4x-2=4*(-2)-2=-10
=>x=-2; y=-10/3
b:\(A=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\left(x+16\right)^2}{\left(x-16\right)\left(\sqrt{x}+2\right)}\)
\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
\(=\sqrt{a-2+2\sqrt{a-2}.2+4}+\sqrt{a-2-2\sqrt{a-2}.2+4}\)
\(=\sqrt{\left(\sqrt{a-2}\right)^2+2\sqrt{a-2}.2+2^2}+\sqrt{\left(\sqrt{a-2}\right)^2-2\sqrt{a-2}.2+2^2}\)
\(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)
Ta có:
\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
\(A=\sqrt{\left(\sqrt{a-2}\right)^2+2.2\sqrt{a-2}+2^2}+\sqrt{\left(\sqrt{a-2}\right)^2-2.2\sqrt{a-2}+2^2}\)
\(A=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(A=|\sqrt{a-2}+2|+|\sqrt{a-2}-2|\) (1)
=> Điều kiên: a - 2 >= 0 <=> a >= 2
(1) => \(A=\sqrt{a-2}+2+|\sqrt{a-2}-2|\)(Do số hạng đầu luôn lớn hơn 0 nên bỏ trị tuyệt đối)
TH1: \(\sqrt{a-2}-2\ge0\Rightarrow A=\sqrt{a-2}+2+\sqrt{a-2}-2\)
\(\sqrt{a-2}\ge2\Rightarrow A=2\sqrt{a-2}\)
\(a\ge6\Rightarrow A=2\sqrt{a-2}\)
(nhận)
TH2: \(\sqrt{a-2}-2\le0\Rightarrow A=\sqrt{a-2}+2-\sqrt{a-2}+2\)
\(\sqrt{a-2}\le2\Rightarrow A=4\)
\(2\le a\le6\Rightarrow A=4\) (Do đkxđ)
Vậy....