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câu a) rút x theo y thế vào A rồi áp dụng HĐT
b)rút xy thế vào B
c)HĐT
d)rút x theo y thé vào C
rồi dùng BĐT cô-si
e)BĐT chưa dấu giá trị tuyệt đối
2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)
c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)
d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
b4 :
\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)
b5:
\(a,ĐK:x\ge1\)
\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)
\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:x\ge5\)
\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)
\(\Leftrightarrow-5\sqrt{x-5}=2\)
\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)
\(c,ĐK:x>0\)
\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)
\(\Leftrightarrow x+9=6\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
Bài đã chữa:
a, x - 9 (x > 0)
= \(\left(\sqrt{x}\right)^2\) - 32
= \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\)
b, x - 5\(\sqrt{x}\) + 4
= x - \(4\sqrt{x}-\sqrt{x}\) +4
= \(\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)\)
Câu d sai đề kìa bạn