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1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Lời giải:
Điều kiện để $Q$ có nghĩa.
\(x>0; x\neq 1\)
\(Q=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\frac{1}{4}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2.\frac{(\sqrt{x}+1)^2-(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(=\frac{1}{4}\left(\frac{x-1}{\sqrt{x}}\right)^2.\frac{x+1+2\sqrt{x}-(x-2\sqrt{x}+1)}{x-1}\)
\(=\frac{1}{4}.\frac{(x-1)^2}{x}.\frac{4\sqrt{x}}{x-1}\)
\(=\frac{x-1}{\sqrt{x}}\)
b)
\(Q=3\sqrt{x}-3\)
\(\Leftrightarrow \frac{x-1}{\sqrt{x}}=3(\sqrt{x}-1)\)
\(\Leftrightarrow \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}=3(\sqrt{x}-1)\)
\(\Leftrightarrow (\sqrt{x}-1)(\frac{\sqrt{x}+1}{\sqrt{x}}-3)=0\)
Vì \(x\neq 1\Rightarrow \sqrt{x}-1\neq 0\). Do đó:
\(\frac{\sqrt{x}+3}{\sqrt{x}}-3=0\Rightarrow 3=2\sqrt{x}\)
\(\Rightarrow x=\frac{9}{4}\) (thỏa mãn)
ây ông ở trên ông ghi là \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
sao xuống dưới lại thành \(\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
sửa lại đi ông ơi
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
3) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{4x-20}=4\)
\(\Leftrightarrow4x-20=16\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
vậy ...
1)
\(A=\dfrac{\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}\right)^2-2^2}\\ A=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{x^2-2x\sqrt{2}+\left(\sqrt{2}\right)^2}{x^2-\sqrt{2}}\\ B=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{\left(x-\sqrt{2}\right)}{\left(x+\sqrt{2}\right)}\)
\(C=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+\left(\sqrt{5}\right)^2}\\ C=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
\(D=\dfrac{\sqrt{a}-2a}{2\sqrt{a}-1}=\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)}{2\sqrt{a}-1}=\sqrt{a}\)
\(E=\dfrac{x^2-2}{x-\sqrt{2}}=\dfrac{x^2-\left(\sqrt{2}\right)^2}{x-\sqrt{2}}\\ E=\dfrac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=x+\sqrt{2}\)
\(F=\dfrac{\sqrt{x}-3}{x-9}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}\right)^2-3^2}\\ F=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ F=\dfrac{1}{\sqrt{x}+3}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Câu a :
\(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}}{2-\sqrt{x}}\times\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{-4x}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)
Câu b :
\(P=-1\)
\(\Leftrightarrow\) \(\dfrac{4x}{\sqrt{x}-3}=-1\)
\(\Leftrightarrow4x=-\sqrt{x}+3\)
\(\Leftrightarrow4x+\sqrt{x}-3=0\)
\(\Leftrightarrow4x+4\sqrt{x}-3\sqrt{x}-3=0\)
\(\Leftrightarrow4\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\dfrac{9}{16}\end{matrix}\right.\)
Vậy \(x=\dfrac{9}{16}\)
Chúc bạn học tốt !!
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4x+2\sqrt{x}-4}{\sqrt{x}^2-2^2}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(\sqrt{x}-2\right)-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-4-4\sqrt{x}-x+x-2\sqrt{x}-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\right)\)
\(P=\dfrac{-4x\left(\sqrt{x}\left(2-\sqrt{x}\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4x\left(-\sqrt{x}\left(\sqrt{x}-2\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{\sqrt{16x^3}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
Có j bạn xem lại coi có sai xót chỗ nào ko nhé, mk ko chắc là đúng 100% đâu.
cảm ơn c nhiều
a: \(P=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{2}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{1}{4-2\sqrt{x}}\)
\(=1:\left(\dfrac{2\left(\sqrt{x}-2\right)-3x+2\sqrt{x}+4}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=1:\dfrac{2\sqrt{x}-4-3x+2\sqrt{x}+4}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=\dfrac{2\left(x-4\right)}{-3x+4\sqrt{x}}\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}+2}{3x-4\sqrt{x}}\)
b: Để P=20 thì \(\sqrt{x}+2=60x-80\sqrt{x}\)
\(\Leftrightarrow60x-81\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=a\)
Pt sẽ là \(60a^2-81a-2=0\)
\(\text{Δ}=\left(-81\right)^2-4\cdot60\cdot\left(-2\right)=7041>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{81-\sqrt{7041}}{120}\left(loại\right)\\a_2=\dfrac{81+\sqrt{7041}}{120}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\left(\dfrac{81+\sqrt{7041}}{120}\right)^2\)