\(\left(x-1\right)^6=\left(x-1\right)^8\)
b) \(\left(2x+1\right)^4=\left(2x+1\right)^8\)
NH
Những câu hỏi liên quan
Tìm x biết:
a) \(\left|x+2\dfrac{1}{2}\right|=\left|3x+1\right|\)
b) \(\left|2x-6\right|+\left|x+3\right|=8\)
c) \(2.\left|x+2\right|+\left|4-x\right|=11\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
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\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
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\(b,\Rightarrow\left[{}\begin{matrix}6-2x-x-3=8\left(x\le-3\right)\\6-2x+x+3=8\left(-3\le x\le3\right)\\2x-6+x+3=8\left(x>3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\left(ktm\right)\\x=1\left(tm\right)\\x=\dfrac{11}{3}\left(tm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{3}\end{matrix}\right.\)
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Xem thêm câu trả lời
a\(\left(2X-1\right)^6=\left(2X-1\right)^8\)
b \(\left(X-1\right)^{X+2}=\left(X-1\right)^{X+4}\)
a) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)
\(\Leftrightarrow\left(2.x-1\right)^8-\left(2.x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},1\right\}\)
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b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy : \(x\in\left\{0,1,2\right\}\)
Chúc học tốt nhé !!
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\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)\(\sqrt{ }\)\(\left|2x+3\right|+\left|2x-1\right|\)=\(\dfrac{8}{3.\left(x+1\right)^2+2}\)
Tìm x:a) 3xleft(3x-8right)-9x^2+80b)6x-15-xleft(5-2xright)0c) x^3-16x0d) 2x^2+3x-50e) 3x^2-xleft(3x-6right)36f) left(x+2right)^2-left(x-5right)left(x+1right)17g) left(x-4right)^2-xleft(x+6right)9h) 4xleft(x-1000right)-x+10000i) x^2-360j) x^2y-2+x+x^2-2y+xy0k) xleft(x+1right)-left(x-1right).left(2x-3right)0l) 3x^3-27x0
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Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
\(sin^6\left(x\right)cos^2\left(x\right)+sin^2\left(x\right)cos^6\left(x\right)=\frac{1}{8}\left(1-cos^4\left(2x\right)\right)\)
2. tìm x
a) \(\left(x-1\right)^3=8\)
b) \(7^{2x-6}=49\)
c) \(\left(2x-14\right)^7=128\)
d) \(x^4.x^5=5^3.5^6\)
e) \(\left[3.\left(x+2\right):7\right].4=120\)
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
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Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
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Giải phương trình:
1. frac{2x+3}{4}-frac{5x+3}{6}frac{3-4x}{12}
2. frac{3.left(2x+1right)}{4}-1frac{15x-1}{10}
3. frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
4. frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
5. frac{x-4}{5}-frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
6. frac{left(x+2right)left(x+10right)}{3}-frac{left(x+4right)left(x+10right)}{12}frac{left(x-2right)left(x+4right)}{4}
7. frac{left(x+2right)^2}{8}-2left(2x-1right)25+frac{left(x-2right)^2}{8}
8.frac{7x^2-14x-5}{5}frac{left(2x+1right)^...
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Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
tính(rút gọn)
a,\(\left(x+3-\frac{1}{x+3}\right)\left(x+\frac{3}{x+4}\right)\)
b,\(\left(2x-4-\frac{x-12}{3x+4}\right)\left(3x-2-\frac{10}{2x+1}\right)\)
c,\(\left(2x-8-\frac{x+10}{3x+1}\right)\left(x-6-\frac{x-6}{3x+2}\right)\)
d,\(\left(1+\frac{1}{x}\right):\left(1-\frac{1}{x^2}\right)\)
Tìm x a, dfrac{left(x+2right)^2}{2} + dfrac{left(1+2xright)^2}{4} + dfrac{left(1-2xright)^2}{8} – (1 + x)2 0b, dfrac{left(x+1right)^2}{2} - dfrac{left(1-2xright)^2}{3} + dfrac{left(1+2xright)^2}{4} - dfrac{left(5-xright)^2}{6} 0c, (3 + x)3 – 3x2(x + 4) + (x + 2)3 (1 – x)3 – 8
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Tìm x
a, \(\dfrac{\left(x+2\right)^2}{2}\) + \(\dfrac{\left(1+2x\right)^2}{4}\) + \(\dfrac{\left(1-2x\right)^2}{8}\) – (1 + x)2 = 0
b, \(\dfrac{\left(x+1\right)^2}{2}\) - \(\dfrac{\left(1-2x\right)^2}{3}\) + \(\dfrac{\left(1+2x\right)^2}{4}\) - \(\dfrac{\left(5-x\right)^2}{6}\)= 0
c, (3 + x)3 – 3x2(x + 4) + (x + 2)3 = (1 – x)3 – 8
a: ta có: \(\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0\)
\(\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0\)
\(\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0\)
\(\Leftrightarrow8x^2+4x+11=0\)
\(\text{Δ}=4^2-4\cdot8\cdot11=-336< 0\)
Vì Δ<0 nên phương trình vô nghiệm
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b.
PT \(\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0\)
\(\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0\)
\(\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0\)
\(\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0\)
$\Leftrightarrow 5x-\frac{15}{4}=0$
$\Leftrightarrow x=\frac{3}{4}$
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c.
PT $\Leftrightarrow (x^3+9x^2+27x+27)-(3x^3+12x^2)+(x^3+6x^2+12x+8)=(-x^3+3x^2-3x+1)-8$
$\Leftrightarrow 42x+42=0$
$\Leftrightarrow x=-1$
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