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1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)
\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=-1\)
b/
\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)
\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)
\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)
\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)
\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra
c/ ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?
e/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+3}+1\right)^2}=x+4\)
\(\Leftrightarrow\sqrt{x+3}+1=x+4\)
\(\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
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)
8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
a/ ĐKXĐ: \(-\frac{1}{2}\le x\le4\)
\(\sqrt{4-x}=\sqrt{x+1}+\sqrt{2x+1}\)
\(\Leftrightarrow4-x=3x+2+2\sqrt{2x^2+3x+1}\)
\(\Leftrightarrow1-2x=\sqrt{2x^2+3x+1}\) (\(x\le\frac{1}{2}\))
\(\Leftrightarrow4x^2-4x+1=2x^2+3x+1\)
\(\Leftrightarrow2x^2-7x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{7}{2}\left(l\right)\end{matrix}\right.\)
Bài này liên hợp cũng được
b/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{5x+1}^2-\sqrt{5x+1}\left(\sqrt{14x+7}-\sqrt{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\Rightarrow x=-\frac{1}{5}\\\sqrt{5x+1}-\sqrt{14x+7}+\sqrt{2x+3}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x+1}+\sqrt{2x+3}=\sqrt{14x+7}\)
\(\Leftrightarrow7x+4+2\sqrt{10x^2+17x+3}=14x+7\)
\(\Leftrightarrow2\sqrt{10x^2+17x+3}=7x+3\)
\(\Leftrightarrow4\left(10x^2+17x+3\right)=\left(7x+3\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-2x}=a\\\sqrt{2x-1}=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a=1-b\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow a^3+\left(1-a\right)^2=1\)
\(\Leftrightarrow a^3+a^2-2a=0\)
\(\Leftrightarrow a\left(a^2+a-2\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2-2x=0\\2-2x=1\\2-2x=-8\end{matrix}\right.\)
d/ ĐKXĐ: \(x\le\frac{5}{4}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5-4x}=a\\\sqrt[3]{x+7}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\a^2+4b^3=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3-b\\a^2+4b^3=33\end{matrix}\right.\)
\(\Leftrightarrow\left(3-b\right)^2+4b^3=33\)
\(\Leftrightarrow4b^3+b^2-6b-24=0\)
\(\Leftrightarrow\left(b-2\right)\left(4b^2+9b+12\right)=0\)
\(\Rightarrow b=2\Rightarrow\sqrt[3]{x+7}=2\Rightarrow x=1\)
a) \(\sqrt{5x+3}=3x-7\)\(\Leftrightarrow\left\{{}\begin{matrix}5x+3=\left(3x-7\right)^2\\3x-7\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+3=9x^2-42x+49\\x\ge\dfrac{7}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-47x+46=0\\x\ge\dfrac{7}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{47+\sqrt{553}}{18}\\x=\dfrac{47-\sqrt{553}}{18}\end{matrix}\right.\\x\ge\dfrac{7}{3}\end{matrix}\right.\)\(\Leftrightarrow\dfrac{47+\sqrt{553}}{18}\).
b) \(\sqrt{3x^2-2x-1}=3x+1\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-1=\left(3x+1\right)^2\\3x+1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+8x+2=0\\x\ge\dfrac{-1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-1\end{matrix}\right.\\x\ge-\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow x=-\dfrac{1}{3}\).
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a: =>|x|+|x-1|=1
Trường hợp 1: x<0
Pt sẽ là -x+1-x=1
=>x=0(loại)
Trường hợp 2: 0<=x<1
Pt sẽ là x+1-x=1
=>1=1(luôn đúng)
Trường hợp 3: x>=1
Pt sẽ là x+x-1=1
=>2x=2
hay x=1(nhận)
b: \(\Leftrightarrow\left|2x-1\right|=2-\sqrt{3}+2+\sqrt{3}=4\)
=>2x-1=4 hoặc 2x-1=-4
=>2x=5 hoặc 2x=-3
=>x=5/2 hoặc x=-3/2