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NQ
0
VT
4
RV
26 tháng 9 2017
Đặt \(a=\sqrt{2-x^2};b=\sqrt{2-\frac{1}{x^2}};c=x+\frac{1}{x}\)
xet x<0 vt < 2 căn 2<3, vt >4=>loại=>x>0=>c>=2;
ta có a+b=4-c;
a^2+b^2=4-x^2-1/x^2=6-c^2;
\(=>\hept{\begin{cases}2a+2b=8-2c\left(2\right)\\a^2+b^2=6-c^2\left(1\right)\end{cases}}\)
trừ 1 cho 2=>a^2-2a+b^2-2b=-c^2-2-2c=>a^2-2b+1+b^2-2b+1=-c^2+2c-1+1
=>\(\left(a-1\right)^2+\left(b-1\right)^2=-\left(c-1\right)^2+1\)
\(< =>\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=1\)
ta lại có (a-1)^2>=0;(b-1)^2>=0;(c-1)^2>=(2-1)^2=1=>Vế trái>=1=Vế phải, dấu bằng xảy ra<=>
\(\hept{\begin{cases}a=1\\b=1\\c=2\end{cases}< =>x=1}\)
26 tháng 9 2017
Bạn tham khảo nhé:Điều kiện bạn tự tìm nhé
pt\(\Leftrightarrow\sqrt{2-x^2}+x-2+\sqrt{2-\frac{1}{x^2}}+\frac{1}{x}-2=0\)
\(\Leftrightarrow\frac{2-x^2-\left(x-2\right)^2}{\sqrt{2-x^2}-x+2}+\frac{2-\frac{1}{x^2}-\left(\frac{1}{x}-2\right)^2}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}=0\)
\(\Leftrightarrow\frac{-2\left(x^2-2x+1\right)}{\sqrt{2-x^2}-x+2}+\frac{-2\left(\frac{1}{x^2}-\frac{2}{x}+1\right)}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{2-x^2}-x+2}+\frac{\left(\frac{1}{x}-1\right)^2}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(\frac{1}{\sqrt{2-x^2}-x+2}+\frac{\frac{1}{x^2}}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\Leftrightarrow x=1\left(N\right)\\\frac{1}{\sqrt{2-x^2}-x+2}+\frac{1}{x\sqrt{2x^2-1}-x+2x^2}=0\left(1\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow x\sqrt{2x^2-1}-x+2x^2+\sqrt{2-x^2}-x+2=0\)
Nhân 2 vào ta có:
\(\Leftrightarrow2x\sqrt{2x^2-1}-4x+4x^2+4+2\sqrt{2-x^2}=0\)
\(\Leftrightarrow\left(x+\sqrt{2x^2-1}\right)^2+\left(\sqrt{2-x^2}+1\right)^2+2\left(x-1\right)^2=0\left(VN\right)\)
Vậy phương trình có 1 nghiệm duy nhất là \(x=1\)
25 tháng 10 2020
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
NN
9
4 tháng 10 2016
ĐKXĐ: z>0
pt<=> \(\frac{x^3+3x^2\sqrt[3]{3x-2}-12x+\sqrt{x}-\sqrt{x}-8}{x}=0\)
<=> \(x^3+3x^2\sqrt[3]{3x+2}-12x-8=0\)
<=> \(3x^2\sqrt[3]{3x-2}-6x^2+x^3-6x^2+12x-8=0\)
<=> \(3x^2\left(\sqrt[3]{3x-2}-2\right)+\left(x-2\right)^3=0\)
<=> \(3x^2\cdot\frac{3x-2-8}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^3=0\)
<=> \(\left(x-2\right)\left(\frac{9x^2}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^2\right)=0\)
<=> \(x=2\)( vì cái trong ngoặc thứ 2 luôn dương vs mọi x>0)
vậy x=2
NH
0
VT
1
1 tháng 9 2017
\(\frac{\sqrt{x}}{1+\sqrt{1-x}}=x^2-2x+2\Leftrightarrow\frac{\sqrt{x}-1}{1+\sqrt{1-x}}+\frac{1}{1+\sqrt{1-x}}-1=x^2-2x+1\)
\(\Leftrightarrow\frac{x-1}{\left(1+\sqrt{1-x}\right)\left(\sqrt{x}+1\right)}+\frac{-\sqrt{1-x}}{1+\sqrt{1-x}}=\left(1-x\right)^2\)
\(\Leftrightarrow\sqrt{1-x}\left[\left(\sqrt{1-x}\right)^3+\frac{\sqrt{1-x}}{\left(1+\sqrt{1-x}\right)\left(\sqrt{x}+1\right)}+\frac{1}{1+\sqrt{1-x}}\right]=0\)
\(\Leftrightarrow\sqrt{1-x}=0\Leftrightarrow x=1.\)
VT
0
\(\sqrt{12-\frac{12}{x^2}}+\sqrt{x^2-\frac{12}{x^2}}=x^2\)
\(pt\Leftrightarrow\sqrt{12-\frac{12}{x^2}}-3+\sqrt{x^2-\frac{12}{x^2}}-1=x^2-4\)
\(\Leftrightarrow\frac{12-\frac{12}{x^2}-9}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{x^2-\frac{12}{x^2}-1}{\sqrt{x^2-\frac{12}{x^2}}+1}=x^2-4\)
\(\Leftrightarrow\frac{\frac{3x^2-12}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{x^4-x^2-12}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-\left(x^2-4\right)=0\)
\(\Leftrightarrow\frac{\frac{3\left(x-2\right)\left(x+2\right)}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{\left(x-2\right)\left(x+2\right)\left(x^2+3\right)}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(\frac{\frac{3}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{x^2+3}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-1\right)=0\)
SUy ra x=±2
\(\sqrt{x^2}\)+\(\sqrt{x^2+3}\)+\(2x^2\)+3+2\(\sqrt{x^2\left(x^2+3\right)}\)=12
Đặt \(\sqrt{x^2}\)+\(\sqrt{x^2+3}\)=a (a>0)
=> \(2x^2\)+3+2\(\sqrt{x^2\left(x^2+3\right)}\)= \(a^2\)
Chị QA 114 đấy