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a.
\(\Leftrightarrow2x^2\ge3\Leftrightarrow x^2\ge\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\le-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)
b.
\(\Leftrightarrow\left(1-x\right)\left(x-3\right)\ge0\Rightarrow1\le x\le3\)
c.
\(\Leftrightarrow\sqrt{1-3x}\le2-x\Leftrightarrow\left\{{}\begin{matrix}1-3x\ge0\\2-x\ge0\\1-3x\le x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\le2\\x^2-x+3\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{1}{3}\)
ĐK: \(x\ge\dfrac{1}{3}\)
\(2x^2+3x-4=\left(4x-3\right)\sqrt{3x-1}\)
\(\Leftrightarrow16x^2+24x-32=8\left(4x-3\right)\sqrt{3x-1}\)
\(\Leftrightarrow\left(4x-3\right)^2+16\left(3x-1\right)-8\left(4x-3\right)\sqrt{3x-1}=25\)
\(\Leftrightarrow\left(4x-3-4\sqrt{3x-1}\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3-4\sqrt{3x-1}=5\\4x-3-4\sqrt{3x-1}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=x-2\\2\sqrt{3x-1}=2x+1\end{matrix}\right.\)
TH1: \(\sqrt{3x-1}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1=\left(x-2\right)^2\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-7x+6=0\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(2\sqrt{3x-1}=2x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(3x-1\right)=\left(2x+1\right)^2\\2x+1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-8x+5\\x\ge-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\) vô nghiệm
Vậy \(x=6\)
ĐKXĐ: \(x\ge\dfrac{1}{3}\)
Đặt \(\sqrt{3x-1}=t\ge0\Rightarrow3x-1=t^2\)
\(\Rightarrow\left\{{}\begin{matrix}2x^2+3x-4=\left(4x-3\right)t\\3x-1=t^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+3x-4=4tx-3t\\2t^2=6x-2\end{matrix}\right.\)
\(\Leftrightarrow2x^2+2t^2+3x-4=4tx-3t+6x-2\)
\(\Leftrightarrow2\left(x-t\right)^2-3\left(x-t\right)-2=0\)
\(\Leftrightarrow...\)
a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)
\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)
\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)
pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)
suy ra tìm đc x
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^2-2x<>0 và x^2-1>0
=>(x>1 và x<>2) hoặc x<-1
b: ĐKXĐ: x+1>0 và 5-3x>0
=>x>-1 và 3x<5
=>-1<x<5/3
c: DKXĐ: 5x+3>=0 và 3-x>0
=>x>=-3/5 và x<3
=>-3/5<=x<3
d: ĐKXĐ: 4-x^2>0 và 1+x>=0
=>x^2<4 và x>=-1
=>-2<x<2 và x>=-1
=>-1<=x<2
e: ĐKXĐ: 2-3x<>0 và 1-6x>0
=>x<>2/3 và x<1/6
=>x<1/6
1.
ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)
\(\Leftrightarrow x^2+x-1+2\sqrt{x\left(x^2-1\right)}=2x^2-3x-4\)
\(\Leftrightarrow x^2-4x-3-2\sqrt{\left(x^2-x\right)\left(x+1\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2-3b^2-2ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow a=3b\)
\(\Leftrightarrow\sqrt{x^2-x}=3\sqrt{x+1}\)
\(\Leftrightarrow x^2-x=9\left(x+1\right)\)
\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)
2.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:
\(x^3+3\left(x^2-4a^2\right)a=0\)
\(\Leftrightarrow x^3+3ax^2-4a^3=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=x\\2a=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+1\\x^2=4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)
Bạn tham khảo thêm ở link sau:
https://hoc24.vn/cau-hoi/giai-phuong-trinhsqrt3x2-5x1-sqrtx2-2sqrt3leftx2-x-1right-sqrtx2-3x4.167769342831
Ta có
x3 - 2x2 - 3x + 3 = √4−x+√1+x
<=>x3 - 2x2 - 3x + 3 = √4+√1
<=>x3 - 2x2 - 3x + 3 = |2| + |1|
<=>x3 - 2x2 - 3x + 3 = 2+1
<=>x3 - 2x2 - 3x + 3 = 3
<=>x3 - 2x2 - 3x =0
<=> (x-3)(x\(^2\)+x)=0
<=>x-3=0 (do x^2 +x khác 0 )
<=> x=3