giải pt:\(\left(\dfrac{x+1}{x-2^{ }}\right)^2+\left(\dfrac{x+1}{x-3}\right)^{ }=12\cdot\left(\dfrac{x-2}{x-3}\right)^2\)
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Giải phương trình
\(1,\dfrac{x^2-2x-3}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-4x+6}{x-2}+\dfrac{x^2-6x+12}{x-3}\)
\(2,\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot[1+\dfrac{1}{x\cdot\left(x+2\right)}]=\dfrac{31}{16}\left(x\in N\right)\)
1)Giải pt: \(2\cdot\left(x+\dfrac{1}{x}\right)^2+\left(x^2+\dfrac{1}{x^2}\right)^2-\left(x^2+\dfrac{1}{x^2}\right)\cdot\left(x+\dfrac{1}{x}\right)^2=\left(x+2\right)^2\)
2)Giải pt: \(\dfrac{|3-2x|-|x|}{|2+3x|+x-2}=5\)
3)tìm tất cả các cặp số nguyên tố(x,y) là nghiệm của pt: x2 - 2y2 - 1=0
1)\(ĐKXĐ:x\ne0\)
Đặt \(\left(x+\dfrac{1}{x}\right)^2=a\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=a-2\)
\(\Rightarrow VT=2a+\left(a-2\right)^2-\left(a-2\right)a\)
\(=2a+a^2-4a+4-a^2+2a=4\)
\(\Rightarrow\left(x+2\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=-4\end{matrix}\right.\)
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Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
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Giải PT này giùm mk nha : \(10\cdot\left(\dfrac{x-2}{x+1}\right)^2+\left(\dfrac{x+2}{x-1}\right)^2-11\cdot\left(\dfrac{x^2-4}{x^2-1}\right)=0\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)
\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)
=>\(-9x^4+81x^2-36=0\)
=>9x^4-81x^2+36=0
=>x^4-9x^2+4=0
=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)
=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)
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26 tháng 2 2023 lúc 17:12
Giải các pt
a) \(\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2\)
b) \(\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1\)
c) \(2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
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b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
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c.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)
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Giải các pt sau:
a)x^2+dfrac{4x^2}{left(x+2right)^2}12
b) dfrac{x^2}{3}+dfrac{48}{x^2}5left(dfrac{x}{3}+dfrac{4}{x}right)
c) left(dfrac{x}{x-1}right)^2+left(dfrac{x}{x+1}right)^2dfrac{10}{9}
d) left(dfrac{x-1}{x}right)^2+left(dfrac{x-1}{x-2}right)^2dfrac{40}{9}
e) x^2+left(dfrac{x}{x-1}right)^28
g) x^3+dfrac{1}{x^3}6left(x+dfrac{1}{x}right)
f) left(x^2+dfrac{1}{x^2}right)+5left(x+dfrac{1}{x}right)-120
Đọc tiếp
Giải các pt sau:
a)\(x^2+\dfrac{4x^2}{\left(x+2\right)^2}=12\)
b) \(\dfrac{x^2}{3}+\dfrac{48}{x^2}=5\left(\dfrac{x}{3}+\dfrac{4}{x}\right)\)
c) \(\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}\)
d) \(\left(\dfrac{x-1}{x}\right)^2+\left(\dfrac{x-1}{x-2}\right)^2=\dfrac{40}{9}\)
e) \(x^2+\left(\dfrac{x}{x-1}\right)^2=8\)
g) \(x^3+\dfrac{1}{x^3}=6\left(x+\dfrac{1}{x}\right)\)
f) \(\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-12=0\)
1: rút gọn rồi tính
left(-dfrac{72}{40}-dfrac{144}{60}-2dfrac{1}{3}right) : left(dfrac{45}{100}-dfrac{25}{60}+-dfrac{75}{25}right)
2: tìm x: 3cdotleft(4-xright)+left(x+2right)cdotleft(1+2xright)7cdotleft(1+xright)-2xcdotleft(2-xright)
3: tìm x: dfrac{2cdotleft(1+xright)}{3}-dfrac{5cdotleft(2-xright)}{6}1dfrac{1}{3}-dfrac{3cdotleft(2x+3right)}{4}-1dfrac{1}{2}cdotleft(x+1right)
4: cho a 3+3^{2^3}+3^3+3^4+...+3^{360}
Đọc tiếp
1: rút gọn rồi tính
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right)\) : \(\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
2: tìm x: \(3\cdot\left(4-x\right)+\left(x+2\right)\cdot\left(1+2x\right)=7\cdot\left(1+x\right)-2x\cdot\left(2-x\right)\)
3: tìm x: \(\dfrac{2\cdot\left(1+x\right)}{3}-\dfrac{5\cdot\left(2-x\right)}{6}=1\dfrac{1}{3}-\dfrac{3\cdot\left(2x+3\right)}{4}-1\dfrac{1}{2}\cdot\left(x+1\right)\)
4: cho a= \(3+3^{2^3}+3^3+3^4+...+3^{360}\)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
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2 tháng 2 2023 lúc 20:06
Giải phương trình:
a) \(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\)
b) \(\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)
a: =>1+3x-6=-x+3
=>3x-5=-x+3
=>4x=8
=>x=2(loại)
b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
=>3x-9+2x-4=x-1
=>5x-13=x-1
=>4x=12
=>x=3(loại)
c: =>x^2-2x+4+x^3+8=12
=>x^3+x^2-2x=0
=>x(x^2+x-2)=0
=>x(x+2)(x-1)=0
=>x=0 hoặc x=1
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giải pt: \(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
Giải:
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)=\left(x-1\right)\left(x-2\right)+\left(x-2\right)\left(x-3\right)\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)
\(\Leftrightarrow x-4=x-2\)
\(\Leftrightarrow0x=2\)
Vậy ...
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