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a: Đặt \(x^2-4=a\)
Pt sẽ là \(a=3\sqrt{xa}\)
\(\Rightarrow a^2=9xa\)
\(\Leftrightarrow a\left(a-9x\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)
hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)
d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)
Pt sẽ là 2a+b=ab+2
=>(b-2)(1-a)=0
=>b=2 và 1-a
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
\(\sqrt{x\left(x-2\right)}+\sqrt{x\left(x-5\right)}=\sqrt{x\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}\right)=0\)
TH1: x = 0 (nhận)
TH2:
\(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-2\right)+\left(\sqrt{x-5}-1\right)-\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\frac{x-2-4}{\sqrt{x-2}+2}+\frac{x-5-1}{\sqrt{x-5}+1}-\frac{x+3-9}{\sqrt{x+3}+3}=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2}+2}+\frac{1}{\sqrt{x-5}+1}-\frac{1}{\sqrt{x+3}+3}\right)\left(x-6\right)=0\)
Pt \(\frac{1}{\sqrt{x-2}+2}+\frac{1}{\sqrt{x-5}+1}-\frac{1}{\sqrt{x+3}+3}=0\) vô no
=> x - 6 = 0
<=> x = 6 (nhận)
a)\(\sqrt{x^2-2x+1}-\sqrt{x^2-4x+4}=x-3\)
\(\Leftrightarrow\left(\sqrt{x^2-2x+1}-3\right)-\left(\sqrt{x^2-4x+4}-2\right)=x-3-1\)
\(\Leftrightarrow\frac{x^2-2x+1-9}{\sqrt{x^2-2x+1}+3}-\frac{x^2-4x+4-4}{\sqrt{x^2-4x+4}+2}=x-4\)
\(\Leftrightarrow\frac{x^2-2x-8}{\sqrt{x^2-2x+1}+3}-\frac{x^2-4x}{\sqrt{x^2-4x+4}+2}-\left(x-4\right)=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-4\right)}{\sqrt{x^2-2x+1}+3}-\frac{x\left(x-4\right)}{\sqrt{x^2-4x+4}+2}-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+2}{\sqrt{x^2-2x+1}+3}-\frac{x}{\sqrt{x^2-4x+4}+2}-1\right)=0\)
Dễ thấy: \(\frac{x+2}{\sqrt{x^2-2x+1}+3}-\frac{x}{\sqrt{x^2-4x+4}+2}-1< 0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=1\)
\(\Leftrightarrow\left(\sqrt{x^2-6x+9}-\frac{7}{2}\right)-\left(\sqrt{x^2+6x+9}-\frac{5}{2}\right)=0\)
\(\Leftrightarrow\frac{x^2-6x+9-\frac{49}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{x^2+6x+9-\frac{25}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}=0\)
\(\Leftrightarrow\frac{\frac{4x^2-24x-13}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{4x^2+24x+11}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}=0\)
\(\Leftrightarrow\frac{\frac{\left(2x-13\right)\left(2x+1\right)}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{\left(2x+1\right)\left(2x+11\right)}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\frac{\frac{2x-13}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{2x+11}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}\right)=0\)
Dễ thấy: \(\frac{\frac{2x-13}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{2x+11}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}< 0\)
\(\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
c)Áp dụng BĐT CAuchy-Schwarz ta có:
\(P^2=\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\)
\(\le\left(1+1\right)\left(x-2+4-x\right)\)
\(=2\cdot\left(x-2+4-x\right)=2\cdot2=4\)
\(\Rightarrow P^2\le4\Rightarrow P\le2\)
a) Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=\sqrt{2}\left(3+4\cdot2-3\right)\)
\(=8\sqrt{2}\)
b) Ta có: \(\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\)
\(=\sqrt{3}\left(1-\frac{1}{3}\cdot\sqrt{9}+2\cdot\sqrt{169}\right)\)
\(=\sqrt{3}\left(1-1+26\right)\)
\(=26\sqrt{3}\)
c) Ta có: \(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}\)
\(=\sqrt{25}\cdot\sqrt{a}+\sqrt{49}\cdot\sqrt{a}-\sqrt{64}\cdot\sqrt{a}\)
\(=\sqrt{a}\left(5+7-8\right)\)
\(=4\sqrt{a}\)
d) Ta có: \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\)
\(=-\sqrt{6b}\cdot\sqrt{6}-\frac{1}{3}\cdot\sqrt{6b}\cdot\sqrt{9}+\frac{1}{5}\cdot\sqrt{6b}\cdot\sqrt{25}\)
\(=-\sqrt{6b}\left(\sqrt{6}+1-1\right)\)
\(=-\sqrt{6b}\cdot\sqrt{6}=-6\sqrt{b}\)
a) Ta có: \(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}.\sqrt{x-2}=20\)
\(\Leftrightarrow2.\sqrt{x-2}+5\sqrt{x-2}-3.\sqrt{x-2}=20\)
\(\Leftrightarrow4.\sqrt{x-2}=20\)
\(\Leftrightarrow\sqrt{x-2}=5\)
\(\Leftrightarrow x-2=25\)
\(\Leftrightarrow x=27\left(TM\right)\)
Vậy \(S=\left\{27\right\}\)
1: =>|2x-1|=5
=>2x-1=5 hoặc 2x-1=-5
=>2x=6 hoặc 2x=-4
=>x=3 hoặc x=-2
2: \(\Leftrightarrow2\sqrt{x-3}+\dfrac{1}{3}\cdot3\sqrt{x-3}-\sqrt{x-3}=4\)
\(\Leftrightarrow\sqrt{x-3}=2\)
=>x-3=4
hay x=7
5: \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=>x-2=0 hoặc x+2=1
=>x=2 hoặc x=-1
Lời giải:
Ta có:
\(\sqrt{x+4}+2\sqrt{x+1}=\sqrt{x+20}\)
\(\Leftrightarrow (\sqrt{x+4}+2\sqrt{x+1})^2=x+20\)
\(\Leftrightarrow 5x+8+4\sqrt{(x+1)(x+4)}=x+20\)
\(\Leftrightarrow \sqrt{(x+1)(x+4)}=3-x\)
Từ đây ta suy ra \(x\leq 3\)
Bình phương hai vế tiếp tục:
\(\Rightarrow (x+1)(x+4)=(3-x)^2\)
\(\Leftrightarrow x^2+5x+4=x^2-6x+9\Leftrightarrow 11x=5\)
\(\Leftrightarrow x=\frac{5}{11}\) (thử lại thấy thỏa mãn đkđb)
sprt=\(\sqrt{ }\)
√(9x - 18) - 1/2 √(4x - 8) + √(x - 2) = 1
⇔ 3√(x - 2) - √(x - 2) + √(x - 2) = 1
⇔ 3√(x - 2) = 1
⇔ √(x - 2) = 1/3 (1)
ĐKXĐ: x ≥ 2
(1) ⇔ x - 2 = 1/9
⇔ x = 1/9 + 2
⇔ x = 19/9 (nhận)
Vậy x = 19/9