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a/ ĐKXĐ: ...
\(\Leftrightarrow4x^2-4x+1-\left(2x-\sqrt{4x-1}\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\frac{\left(2x-1\right)^2}{2x+\sqrt{4x-1}}=0\)
\(\Leftrightarrow\left(2x-1\right)^2\left(1-\frac{1}{2x+\sqrt{4x-1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\2x+\sqrt{4x-1}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x-1}=1-2x\) (\(x\le\frac{1}{2}\))
\(\Leftrightarrow4x-1=\left(1-2x\right)^2\)
\(\Leftrightarrow4x-1=4x^2-4x+1\)
\(\Leftrightarrow2x^2-4x+1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{2}}{2}\left(l\right)\\x=\frac{2-\sqrt{2}}{2}\end{matrix}\right.\)
b/
Đặt \(3x^2-2x+2=a>0\) ta được:
\(\sqrt{a+7}+\sqrt{a}=7\)
\(\Leftrightarrow2a+7+2\sqrt{a^2+7a}=49\)
\(\Leftrightarrow\sqrt{a^2+7a}=21-a\) (\(a\le21\))
\(\Leftrightarrow a^2+7a=\left(21-a\right)^2\)
\(\Leftrightarrow a^2+7a=a^2-42a+441\)
\(\Rightarrow a=9\Rightarrow3x^2-2x+2=9\)
\(\Leftrightarrow3x^2-2x-7=0\Rightarrow x=\frac{1\pm\sqrt{22}}{3}\)
a)
Đkxđ: \(\left\{{}\begin{matrix}-3x+2\ge0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{-2}{3}\\x\ne-1\end{matrix}\right.\)
b)
Đkxđ: \(\left\{{}\begin{matrix}x-2\ge0\\-x-4\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le4\end{matrix}\right.\)\(\Leftrightarrow2\le x\le4\).
c)
Đkxđ: \(\left\{{}\begin{matrix}3x^2+6x+11>0\\2x+1\ge0\end{matrix}\right.\)\(\Leftrightarrow2x+1\ge0\)\(\Leftrightarrow x\ge-\dfrac{1}{2}\).
d)
Đkxđ: \(\left\{{}\begin{matrix}x+4\ge0\\x^2-9\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\ne3\\x\ne-3\end{matrix}\right.\).
a/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
Ta được:
\(1+\frac{a^2-1}{3}=a\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x\left(1-x\right)}=0\\2\sqrt{x-x^2}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\left(1-x\right)=0\\-4x^2+4x-9=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\sqrt{x+5}=a\ge0\Rightarrow a^2-x=5\)
\(x^2+a=a^2-x\)
\(\Leftrightarrow x^2-a^2+a+x=0\)
\(\Leftrightarrow\left(a+x\right)\left(x-a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-x\left(x\le0\right)\\\sqrt{x+5}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2\left(x\le0\right)\\x+5=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
c/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
ĐKXĐ: ...
\(2x+1-\sqrt{3x^2+7x}+x-\sqrt{3x-1}=0\)
\(\Leftrightarrow\frac{4x^2+4x+1-\left(3x^2+7x\right)}{2x+1+\sqrt{3x^2+7x}}+\frac{x^2-\left(3x-1\right)}{x+\sqrt{3x-1}}=0\)
\(\Leftrightarrow\frac{x^2-3x+1}{2x+1+\sqrt{3x^2+7x}}+\frac{x^2-3x+1}{x+\sqrt{3x-1}}=0\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left(\frac{1}{2x+1+\sqrt{3x^2+7x}}+\frac{1}{x+\sqrt{3x-1}}\right)=0\)
\(\Leftrightarrow x^2-3x+1=0\Rightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow a=3;b=5;c=2\)
a)
Pt\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=\left(x-3\right)^2\\x-3\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4=x^2-6x+9\\x\ge3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+13=0\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{9+\sqrt{29}}{2}\\x_2=\dfrac{9-\sqrt{29}}{2}\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{9+\sqrt{29}}{2}\)
Vậy \(x=\dfrac{9+\sqrt{29}}{2}\) là nghiệm của phương trình.
b) Pt \(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+3=\left(2x-1\right)^2\\2x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-2=0\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{7}}{3}\\x_2=\dfrac{1-\sqrt{7}}{3}\end{matrix}\right.\\x\ge\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{7}}{3}\)
Vậy phương trình có duy nhất nghiệm là: \(x=\dfrac{1+\sqrt{7}}{3}\)
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: \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\\x\ne\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Đặt \(x^2-x-1=a\) ta được:
\(\frac{4}{a-1}+\frac{2}{a}=5\Leftrightarrow4a+2\left(a-1\right)=5a\left(a-1\right)\)
\(\Leftrightarrow5a^2-11a+2=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x-1=2\\x^2-x-1=\frac{1}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\5x^2-5x-6=0\end{matrix}\right.\) (bấm máy)
b/ ĐKXĐ: \(x>2\)
Đặt \(\sqrt{x-2}=a>0\)
\(\frac{4}{a+1}-\frac{1}{a}=1\Leftrightarrow4a-\left(a+1\right)=a\left(a+1\right)\)
\(\Leftrightarrow a^2-2a+1=0\Rightarrow a=1\)
\(\Rightarrow\sqrt{x-2}=1\Rightarrow x=3\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)
\(\Leftrightarrow4\left(2-3\sqrt{x}\right)-\left(\sqrt{x}+1\right)=3\left(\sqrt{x}+1\right)\left(2-3\sqrt{x}\right)\)
\(\Leftrightarrow9x-10\sqrt{x}+1=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=\frac{1}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{81}\end{matrix}\right.\)
ĐKXĐ: ...
\(3x\left(3x-\sqrt{8x^2+x+5}\right)-2\left(\sqrt{x^2-x-1}-2\right)=0\)
\(\Leftrightarrow\frac{3x\left(x^2-x-5\right)}{3x+\sqrt{8x^2+x+5}}-\frac{2\left(x^2-x-5\right)}{\sqrt{x^2-x-5}+2}=0\)
\(\Leftrightarrow\left(x^2-x-5\right)\left(\frac{3x}{3x+\sqrt{8x^2+x+5}}-\frac{2}{\sqrt{x^2-x-5}+2}\right)=0\)
May mắn là người ta chỉ bắt tìm 1 nghiệm nên cái ngoặc kia ta khỏi quan tâm, nếu không thì cách liên hợp này không ổn đâu
\(x^2-x-5=0\Rightarrow x=\frac{1+\sqrt{21}}{2}\) \(\Rightarrow a=1;b=21;c=2\)
Nguyễn Việt Lâm rep ib mk vss