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a/ ĐKXĐ: \(x^2+2x-6\ge0\)
\(\Leftrightarrow x^2+2x-6+\left(x-2\right)\sqrt{x^2+2x-6}=0\)
\(\Leftrightarrow\sqrt{x^2+2x-6}\left(\sqrt{x^2+2x-6}+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-6}=0\left(1\right)\\\sqrt{x^2+2x-6}=2-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+2x-6=0\Rightarrow x=-1\pm\sqrt{7}\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\x^2+2x-6=\left(2-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\6x=10\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{3}\)
Câu b nhìn ko ra hướng, ko biết đề có nhầm đâu ko :(
c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(x^2+x\right)\left(x^2+x+2\right)}-\left(3-x\right)\sqrt{x^2+x}=0\)
\(\Leftrightarrow\sqrt{x^2+x}\left(\sqrt{x^2+x+2}-3+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\left(1\right)\\\sqrt{x^2+x+2}=3-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)
d/
Ta có \(\sqrt{x^2+3x+4}=\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{7}{4}}>1\)
\(\Rightarrow\sqrt{x^2+3x+4}-1>0\)
Nhân 2 vế của pt với \(\sqrt{x^2+3x+4}-1\)
\(\left(\sqrt{x^2+3x+4}-1\right)\left(x^2+3x+3\right)=3x\left(x^2+3x+3\right)\)
\(\Leftrightarrow\left(x^2+3x+3\right)\left(\sqrt{x^2+3x+4}-1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+3=0\left(vn\right)\\\sqrt{x^2+3x+4}=3x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)
a/ ĐKXĐ: \(0\le x\le4\)
\(\left(x^2-4x\right)\sqrt{-x^2+4x}+x^2-4x+2=0\)
Đặt \(\sqrt{-x^2+4x}=a\ge0\)
\(-a^2.a-a^2+2=0\)
\(\Leftrightarrow a^3+a^2-2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+2a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a^2+2a+2=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-x^2+4x}=1\Leftrightarrow x^2-4x+1=0\Rightarrow...\)
b/ \(x^4+2x^2+x\sqrt{2x^2+4}-4=0\)
Đặt \(x\sqrt{2x^2+4}=a\Rightarrow x^2\left(2x^2+4\right)=a^2\Rightarrow x^4+2x^2=\frac{a^2}{2}\)
\(\frac{a^2}{2}+a-4=0\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=2\left(x>0\right)\\x\sqrt{2x^2+4}=-4\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^4+4x^2=4\\2x^4+4x^2=16\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1\\x^2=-\sqrt{3}-1\left(l\right)\\x^2=2\\x^2=-4\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)
c/ Đặt \(\sqrt[3]{2x^2+3x-10}=a\Rightarrow2x^2+3x=a^3+10\)
\(a^3+10-14=2a\)
\(\Leftrightarrow a^3-2a-4=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^2+2a+2\right)=0\Rightarrow a=2\)
\(\Rightarrow\sqrt[3]{2x^2+3x-10}=2\Rightarrow2x^2+3x-18=0\Rightarrow...\)
d/ \(\Leftrightarrow2\left(3x^2+x+4\right)+\sqrt[3]{3x^2+x+4}-18=0\)
Đặt \(\sqrt[3]{3x^2+x+4}=a\)
\(2a^3+a-18=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a^2+4a+9\right)=0\Rightarrow a=2\)
\(\Rightarrow\sqrt[3]{3x^2+x+4}=2\Rightarrow3x^2+x-4=0\Rightarrow...\)
e/ \(\Leftrightarrow x^2+5x+2-3\sqrt{x^2+5x+2}-2=0\)
Đặt \(\sqrt{x^2+5x+2}=a\ge0\)
\(a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=\frac{3+\sqrt{17}}{2}\\a=\frac{3-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+2}=\frac{3+\sqrt{17}}{2}\Rightarrow x^2+5x-\frac{9+3\sqrt{17}}{2}=0\)
Bài cuối xấu quá, chắc nhầm số liệu
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)
a/ ĐKXĐ: \(x^2+3x+2\ge0\)
\(\Leftrightarrow3-2\sqrt{x^2+3x+2}=1-2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\sqrt{x^2+3x+2}=\sqrt{x^2-x+1}+1\)
\(\Leftrightarrow x^2+3x+2=x^2-x+1+1+2\sqrt{x^2-x+1}\)
\(\Leftrightarrow2x=\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=x^2-x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3x^2+x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{-1+\sqrt{13}}{6}\\x=\frac{-1-\sqrt{13}}{6}\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(3x^2-7x+2\ge0\)
\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\) (1)
\(\Rightarrow3x^2-5x+7=9+3x^2-7x+2-6\sqrt{3x^2-7x+2}\)
\(\Rightarrow2-x=3\sqrt{3x^2-7x+2}\) (\(x\le2\))
\(\Rightarrow\left(2-x\right)^2=9\left(3x^2-7x+2\right)\)
\(\Rightarrow x^2-4x+4=27x^2-63x+18\)
\(\Rightarrow26x^2-59x+14=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)
Do bước biến đổi thứ 2 ko phải phép tương đương nên cần thay 2 nghiệm vào (1) để kiểm tra lại, bạn tự thay nhé
a/ \(\Leftrightarrow x^2+5x-2-2\sqrt[3]{x^2+5x-2}+4=0\)
Đặt \(\sqrt[3]{x^2+5x-2}=a\)
\(a^3-2a+4=0\)
\(\Leftrightarrow\left(a+2\right)\left(a^2-2a+2\right)=0\Rightarrow a=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\Rightarrow x^2+5x+6=0\Rightarrow...\)
b/ ĐKXĐ:...
\(\Leftrightarrow-3\left(-x^2+4x+10\right)-5\sqrt{-x^2+4x+10}+42=0\)
Đặt \(\sqrt{-x^2+4x+10}=a\ge0\)
\(-3a^2-5a+42=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{14}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+4x+10}=3\Rightarrow x^2-4x-1=0\Rightarrow...\)
c/ ĐKXĐ: ...
\(\Leftrightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\)
Đặt \(\sqrt{x^2+3x}=a\ge0\)
\(a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+3x}=2\Rightarrow x^2+3x-4=0\)
d/ ĐKXĐ: \(-1\le x\le2\)
\(\Leftrightarrow\sqrt{3-x+x^2}=1+\sqrt{2+x-x^2}\)
\(\Leftrightarrow3-x+x^2=3+x-x^2+2\sqrt{2+x-x^2}\)
\(\Leftrightarrow2+x-x^2+\sqrt{2+x-x^2}-2=0\)
Đặt \(\sqrt{2+x-x^2}=a\ge0\)
\(a^2+a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2+x-x^2}=1\Leftrightarrow x^2-x-1=0\)
e/ \(\Leftrightarrow\sqrt{x^2-3x+3}-1+\sqrt{x^2-3x+6}-2=0\)
\(\Leftrightarrow\frac{x^2-3x+2}{\sqrt{x^2-3x+3}+1}+\frac{x^2-3x+2}{\sqrt{x^2-3x+6}+2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{\sqrt{x^2-3x+3}+1}+\frac{1}{\sqrt{x^2-3x+6}+2}\right)=0\)
\(\Leftrightarrow x^2-3x+2=0\)
a/ ĐKXĐ: ...
\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)
\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))
\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-3\)
\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)
\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\le12\)
\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)
\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)
\(\Leftrightarrow a^3+a^2-12a=0\)
\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)
ĐKXĐ:2<x<5
\(\sqrt{x^2-7x+10}=3x-1(x\ge \frac{1}{3}) \)
<=>\(x^2-7x+10=9x^2-6x+1 \)
<=>\(8x^2+x-9=0\)
<=>\((x-1)(8x+9)=0 \)
<=>x=1(x\(\ge\frac{1}{3} \))
b,
\(\left|3x-5\right|=2x^2+x-3\)(1)
Nếu \(3x-5\ge0\Leftrightarrow x\ge\dfrac{5}{3}\)
Thì pt (1) <=> \(3x-5=2x^2+x-3\)
\(\Leftrightarrow2x^2-2x+2=0\)
\(\Leftrightarrow PT\) vô \(n_o\) (vì \(\Delta< 0\))
Nếu 3x - 5 <0 \(\Leftrightarrow x< \dfrac{5}{3}\)
Thì pt (1) <=> \(5-3x=2x^2+x-3\)
\(\Leftrightarrow2x^2+4x-8=0\)
\(\Leftrightarrow x^2+2x-4=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)(t/m)
Vậy....