Giải phương trình 2 x + 3 + 4 x 2 + 9 x + 2 = 2 x + 2 + 4 x + 1 .
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\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\)
\(ĐKXĐ:x\ne\pm2\)
\(pt\Leftrightarrow\frac{9}{x^2-4}=\frac{x^2-3x+2}{x^2-4}+\frac{3x+6}{x^2-4}\)
\(\Leftrightarrow\frac{9}{x^2-4}=\frac{x^2+8}{x^2-4}\)
\(\Leftrightarrow x^2+8=9\Leftrightarrow x=\pm1\left(tm\right)\)
Vậy pt có 2 nghiệm là 1 và -1
Điều kện : \(x+2\ne0\) và \(x-2\ne0\Leftrightarrow x=\pm2\)
( Khi đó \(x^2-4=\left(x+2\right)\left(x-2\right)\ne0\) )
\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\)
\(\Leftrightarrow\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)+3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-3x+2+3x+6=9\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)
Vậy tập nghiệm của PT là: \(S=\left\{-1;1\right\}\)
Chúc bạn học tốt !!!
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a) Áp dụng bđt AM-GM có:
\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)
\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)
Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)
Vậy...
b)Đk:\(x\ge2\)
Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)
Do \(x\ge2\Rightarrow x-1>0\)
Chia cả hai vế của pt cho x-1 ta được:
\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy S={2}
c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)
Thay x=3 vào pt thấy thỏa mãn
Vậy S={3}
a) Quên mất, ko áp dụng đc AM-GM, xin lỗi
Pt \(\Leftrightarrow\sqrt[3]{9-x}-2=2-\sqrt[3]{7+x}\)
\(\Leftrightarrow\dfrac{9-x-8}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{8-\left(7-x\right)}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)
\(\Leftrightarrow\dfrac{1-x}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1-x}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4=4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}\left(1\right)\end{matrix}\right.\)
Từ (1) \(\Leftrightarrow\sqrt[3]{\left(9-x\right)^2}-\sqrt[3]{\left(7+x\right)^2}+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)\left(\sqrt[3]{9-x}+\sqrt[3]{7+x}\right)+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right).4+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\sqrt[3]{9-x}-\sqrt[3]{7+x}=0\)
\(\Leftrightarrow\sqrt[3]{9-x}=\sqrt[3]{7+x}\)\(\Leftrightarrow9-x=7+x\)
\(\Leftrightarrow x=1\)
Vậy S={1}
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a: \(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
=>(4x+14+3x+9)(4x+14-3x-9)=0
=>(7x+23)(x+5)=0
=>x=-23/7 hoặc x=-5
\(a,\\ \Leftrightarrow7x^2+58x+115=0\\ \Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
\(b,\\ \Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=0\\ \LeftrightarrowĐặt.x^2+6x+5=a\\ \Leftrightarrow a=a\left(a+3\right)=10\\ \Leftrightarrow a^2+3a-10=0\\ \Leftrightarrow\left(a+5\right)\left(a-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-5\\a=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+6x+5=-5\\x^2+6x+5=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+6x+10=0\\x^2+6x+3=0\end{matrix}\right.\\ \left(Vô.n_o\Delta=36-40=-4< 0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{6}\\x=-3-\sqrt{6}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐK: \(x\le-1,x\ge3\)
\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)
\(\Leftrightarrow x^2-2x-3=1\)
\(\Leftrightarrow x^2-2x-4=0\)
\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)
b, ĐK: \(-2\le x\le2\)
Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)
Khi đó phương trình tương đương:
\(3t-t^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)
Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)
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Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x^2}{9-x^2}\) ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{x^2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{3x^2-7x^2}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2=3x^2-7x^2\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-3x^2-x^2-3x^2-3x^2+7x^2\right)-3x=0\)
\(\Leftrightarrow-3x^2-3x=0\)
\(\Leftrightarrow-3x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
KL : nghiệm của PT là : \(S=\left\{0;-1\right\}\)
\(\frac{x-4}{x-1}+\frac{x+4}{x+1}=2\) DKXĐ : \(x\ne\pm1\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=2\)
\(\Leftrightarrow x^2+x-4x-4+x^2-x+4x-4=2\)
\(\Leftrightarrow\left(x^2+x^2\right)\left(x-4x-x+4x\right)+\left(-4-4\right)=2\)
\(\Leftrightarrow2x^2-8=2\)
\(\Leftrightarrow2x^2=10\)
.....
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a: \(\Leftrightarrow x^2-2x+1-x^2-2x-1=2x-6\)
=>2x-6=-4x
=>6x=6
hay x=1
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\)
=>(x-3)(-4x+1)=0
=>x=3 hoặc x=1/4
c: \(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)-x^2+4x-4=0\)
\(\Leftrightarrow3x^2+16x+5-3x^2+48=0\)
=>16x+53=0
hay x=-53/16
d: \(\Leftrightarrow x^3+4x^2-9x-36=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-9\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
b)x^2-9=(x-3)(5x+2)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(1-4x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\1-4x=0\end{matrix}\right.\left\{{}\begin{matrix}x=0+3\\x=1:4\end{matrix}\right.\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
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1) `x^2+4-2(x-1)=(x-2)^2`
`<=>x^2+4-2x+2=x^2-4x+4`
`<=>-2x+2=-4x`
`<=>2x=-2`
`<=>x=-1`
.
2) ĐKXĐ: `x \ne \pm 3`
`(x+3)/(x-3)-(x-1)/(x+3)=(x^2+4x+6)/(x^2-9)`
`<=>(x+3)^2-(x-1)(x-3)=x^2+4x+6`
`<=>x^2+6x+9-x^2+4x-3=x^2+4x+6`
`<=>10x+6=x^2+4x+6`
`<=>x^2-6x=0`
`<=>x(x-6)=0`
`<=>x=0;x=6`
.
3) ĐKXĐ: `x \ne \pm 3`
`(3x-3)/(x^2-9) -1/(x-3 )= (x+1)/(x+3)`
`<=>(3x-3)-(x+3)=(x+1)(x-3)`
`<=> 2x-6=x^2-2x-3`
`<=>x^2-4x+3=0`
`<=>x^2-x-3x+3=0`
`<=>x(x-1)-3(x-1)=0`
`<=>(x-3)(x-1)=0`
`<=> x=3;x=1`
Vậy...
Pt ⇔ 2 x + 3 + ( x + 2 ) ( 4 x + 1 ) = 2 x + 2 + 4 x + 1 . ĐK: x ≥ − 1 4
Đặt t 2 = 8 x + 4 ( x + 2 ) ( 4 x + 1 ) + 9 ⇔ 2 x + ( x + 2 ) ( 4 x + 1 ) = t 2 − 9 4
PTTT t 2 − 4 t + 3 = 0 ⇔ t = 1 hoặc t = 3
TH1. t = 1 giải ra vô nghiệm hoặc kết hợp với ĐK t ≥ 7 bị loại
TH 2 t = 3 ⇒ 2 x + 2 + 4 x + 1 = 3. Giải pt tìm được x = − 2 9 (TM)
Vậy pt có nghiệm duy nhất x = − 2 9
\(x=1,0572\)