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1/ \(\left\{{}\begin{matrix}10-9x\ge0\\3x-2=\left(10-9x\right)^2\end{matrix}\right.< =>\left\{{}\begin{matrix}x\le\frac{10}{9}\\3x-2=100-180x+81x^2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\le\frac{10}{9}\\\left[{}\begin{matrix}x=\frac{34}{27}\left(l\right)\\x=1\end{matrix}\right.\end{matrix}\right.\)=> x=1
Vậy ....
2.\(\left\{{}\begin{matrix}x\ge3\\3x-4=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\3x-4=x^2-6x+9\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\ge3&x=\frac{9\pm\sqrt{29}}{2}&\end{matrix}\right.\)=> \(x=\frac{9+\sqrt{29}}{2}\)
Vậy ....
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
1: =>3x^2+5x-7=3x+14
=>2x=21
=>x=21/2
2;=>x+4=4
=>x=0
3: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-20x+25-4x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-24x+32=0\end{matrix}\right.\)
=>x>=5/2 và x^2-6x+8=0
=>x=4
4: \(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2+2x-1=x^2-2x+1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
5: \(\Leftrightarrow\sqrt{2x+16}=x-4\)
=>x>=4 và x^2-8x+16=2x+16
=>x>=4 và x^2-10x=0
=>x=10
đa phần mình sử dụng phương pháp liên hợp nha bạn
\(\sqrt{a}-\sqrt{b}=\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
b. điều kiện \(\dfrac{1}{4}\le x\le\dfrac{3}{8}\), pt:
\(\Leftrightarrow\sqrt{3-8x}-\sqrt{4x-1}=6x-2\\ \Leftrightarrow\dfrac{3-8x-4x+1}{\sqrt{3-8x}+\sqrt{4x-1}}=2\left(3x-1\right)\\ \Leftrightarrow\dfrac{-4\left(3x-1\right)}{\sqrt{3-8x}+\sqrt{4x-1}}=2\left(3x-1\right)\\ \Leftrightarrow2\left(3x-1\right)+\dfrac{4\left(3x-1\right)}{\sqrt{3-8x}+\sqrt{4x-1}}=0\\ \Leftrightarrow2\left(3x-1\right)\left(1+\dfrac{2}{\sqrt{3-8x}+\sqrt{4x-1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(n\right)\\1+\dfrac{2}{\sqrt{3-8x}+\sqrt{4x-1}}=0\left(vn\right)\end{matrix}\right.\)
d. điều kiện: \(x\le-4\cup x\ge0\), pt:
\(\Leftrightarrow1-\sqrt{x^2-3x+3}=\sqrt{2x^2+x+2}-\sqrt{x^2+4x}\\ \Leftrightarrow\dfrac{1-x^2+3x-3}{1+\sqrt{x^2-3x+3}}=\dfrac{2x^2+x+2-x^2-4x}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\\ \Leftrightarrow\dfrac{-\left(x-1\right)\left(x-2\right)}{1+\sqrt{x^2-3x+3}}=\dfrac{\left(x-1\right)\left(x-2\right)}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=1\left(n\right)\\\dfrac{-1}{1+\sqrt{x^2-3x+3}}=\dfrac{1}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\left(vn\right)\end{matrix}\right.\)
e. điều kiện:x thuộc R
\(\Leftrightarrow\sqrt{x^2+15}-4=3x-3+\sqrt{x^2+8}-3\\ \Leftrightarrow\dfrac{x^2+15-16}{\sqrt{x^2+15}+4}=3\left(x-1\right)+\dfrac{x^2+8-9}{\sqrt{x^2+8}+3}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+15}+4}-3\left(x-1\right)-\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+8}+3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{\left(x+1\right)}{\sqrt{x^2+15}+4}-3-\dfrac{\left(x+1\right)}{\sqrt{x^2+8}+3}=0\left(1\right)\end{matrix}\right.\)
(1) mình không biết có vô nghiệm không nữa và cũng thua luôn
f. điều kiện: \(x\ge-2\)
bài này giải cách hơi khác một chút
đặt \(a=\sqrt{x+5}\left(\ge0\right)\\ b=\sqrt{x+2}\left(\ge0\right)\)
pt:
\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left[\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)\right]\\ \Rightarrow\left(a-b\right)\left(1+ab\right)=3\left(1\right)\)
mà \(a^2-b^2=x+5-x-2=3\\ \Rightarrow\left(a-b\right)\left(a+b\right)=3\left(2\right)\)
=> (1) = (2)
\(\Leftrightarrow\left(a-b\right)\left(1+ab\right)=\left(a-b\right)\left(a+b\right)\\ \Leftrightarrow\left(a-b\right)\left(1+ab-a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)
TH1: a=b \(\Leftrightarrow\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow x+5=x+2\left(vn\right)\)
TH2: a=1\(\Leftrightarrow\sqrt{x+5}=1\Leftrightarrow x=-4\left(l\right)\)
TH3: b=1\(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\left(n\right)\)
g. điều kiện: \(x\le-\sqrt{2}\cup x\ge\dfrac{7+\sqrt{37}}{2}\)
pt:
\(\dfrac{3x^2-7x+3-3x^2+5x+1}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{x^2-2-x^2+3x-4}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\\ \Leftrightarrow\dfrac{-2\left(x-2\right)}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{3\left(x-2\right)}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(n\right)\\\dfrac{-2}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{3}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\left(vn\right)\end{matrix}\right.\)h. điều kiện \(x\le-2-\sqrt{7}\cup x\ge-2+\sqrt{7}\)
\(\sqrt{2x^2+x-1}-\sqrt{x^2+4x-3}=\sqrt{2x^2+4x-3}-\sqrt{3x^2+x-1}\\ \Leftrightarrow\dfrac{2x^2+x-1-x^2-4x+3}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{2x^2+4x-3-3x^2-x+1}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\\ \Leftrightarrow\dfrac{x^2-3x+2}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{-\left(x^2-3x+2\right)}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\Leftrightarrow x=1\left(n\right),x=2\left(n\right)\\\dfrac{1}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{-1}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\left(vn\right)\end{matrix}\right.\)
(nhớ tích cho mình nha, mấy bài kia mình ko biết làm huhu)
Đệ biết là có người làm câu c,d nên xin xí câu e :3
ĐK: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)
\(PT\Leftrightarrow5+\sqrt{x+1}=7\left(x-2\right)\)
\(\Leftrightarrow\sqrt{x+1}=7x-19\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\x+1=49x^2-266x+361\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\49x^2-267x+360=0\end{matrix}\right.\)
\(\Rightarrow x=3\left(tm\right)\)
a/ \(\Leftrightarrow\left\{{}\begin{matrix}9-2x\ge0\\x^2-4x-12=\left(9-2x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{9}{2}\\3x^2-32x+93=0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ \(\Leftrightarrow\left(x+1\right)\sqrt[3]{15x^2-x-1}-\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\sqrt[3]{15x^2-x-1}-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\\sqrt[3]{15x^2-x-1}-x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{15x^2-x-1}=x-1\)
\(\Leftrightarrow15x^2-x-1=x^3-3x^2+3x-1\)
\(\Leftrightarrow x^3-18x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-18x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=9\pm\sqrt{77}\\\end{matrix}\right.\)
Viết đề mà ko ai đọc được vậy :v
a) \(3x^2+2x+3=\left(3x+1\right)\sqrt{x^2+3}\)
\(\Leftrightarrow3x^2+2x+3-3x\sqrt{x^2+3}-\sqrt{x^2+3}=0\)
\(\Leftrightarrow x^2+3-x\sqrt{x^2+3}-\sqrt{x^2+3}-2x\sqrt{x^2+3}+2x^2+2x=0\)
\(\Leftrightarrow\sqrt{x^2+3}\cdot\left(\sqrt{x^2+3}-x-1\right)-2x\cdot\left(\sqrt{x^2+3}-x-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+3}-x-1\right)\left(\sqrt{x^2+3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\) ( thỏa mãn )
Vậy...
\(\left(4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\) (1)
<=>\(\left(4x-1\right)\left[\sqrt{x^2+1}-\left(3-x\right)\right]=6x^2-11x+4\)
Xét \(\sqrt{x^2+1}+3-x=0\)
<=> \(x^2+1=x^2-6x+9\) <=>\(x=\frac{4}{3}\)(tm phương trình (1))
Xét \(\sqrt{x^2+1}+3-x\ne0\)
pt <=>\(\frac{\left(4x-1\right)\left(x^2+1-x^2+6x-9\right)}{\sqrt{x^2+1}+3-x}=\left(3x-4\right)\left(2x-1\right)\)
<=> \(\frac{\left(4x-1\right)\left(6x-8\right)}{\sqrt{x^2+1}+3-x}-\left(3x-4\right)\left(2x-1\right)=0\)
<=>\(\left(3x-4\right)\left(\frac{2\left(4x-1\right)}{\sqrt{x^2+1}+3-x}-2x+1\right)=0\)
<=>\(\left[{}\begin{matrix}x=\frac{4}{3}\left(tm\right)\\\frac{8x-2}{\sqrt{x^2+1}+3-x}-2x+1=0\left(2\right)\end{matrix}\right.\)
pt (2) <=>\(8x-2=\left(2x-1\right)\sqrt{x^2+1}-2x^2+7x-3\)
<=>\(2x^2+x+1=\left(2x-1\right)\sqrt{x^2+1}\)( đk: \(x\ge\frac{1}{2}\))
=>\(4x^4+x^2+1+4x^3+2x+4x^2=\left(2x-1\right)^2\left(x^2+1\right)\)
<=>\(4x^4+4x^3+5x^2+2x+1=4x^4-4x^3+5x^2-4x+1\)
<=>\(8x^3+6x=0\) <=> \(x\left(8x^2+6\right)=0\) <=>x=0 (do 8x2+6>0) (không t/m (2))
=>(2) vô nghiệm
Vậy pt có tập nghiệm \(S=\left\{\frac{4}{3}\right\}\)
P/s: Hơi dài :)