Tìm x , biết :
\(a,\text{ }2^x\cdot4=128\)
\(b,\text{ }x^{15}=x\)
\(c,\text{ }\left(2x+1\right)^3=125\)
\(d,\text{ }\left(x-5\right)^4=\left(x-5\right)^6\)
\(e,\text{ }\left(2x-15\right)^5=\left(2x-15\right)^3\)
Những câu hỏi liên quan
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
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12 tháng 5 2021 lúc 8:38
text{Tìm x, biết:}a) 20text{%}x-x+dfrac{1}{5}dfrac{3}{4}b) dfrac{2x+1}{3}dfrac{x-5}{2}c) left(x-dfrac{3}{4}right)left(4+3xright)0d) x-dfrac{1}{3}x+dfrac{1}{5}xdfrac{-26}{5}e) 50text{%}x+dfrac{2}{3}xx-5g) dfrac{2}{3}left(x+dfrac{9}{5}right)-dfrac{3}{10}.left(5x-dfrac{1}{3}right)dfrac{7}{15}
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\(\text{Tìm x, biết:}\)
\(a\)) \(20\text{%}x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(b\)) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(c\)) \(\left(x-\dfrac{3}{4}\right)\left(4+3x\right)=0\)
\(d\)) \(x-\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{-26}{5}\)
\(e\)) \(50\text{%}x+\dfrac{2}{3}x=x-5\)
\(g\)) \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}.\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
câu c) mang tính mua vui hay gì hả bn
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mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
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Xem thêm câu trả lời
Giải bất phương trình:
\(\text{ a) }\left|3x-2\right|< 4\)
\(\text{b) }\left|3-2x\right|< x+1\)
\(\text{c) }\left|3x-1\right|>5\)
\(\text{d) }\left|x+1\right|>\left|x-2\right|\)
\(\text{e) }\left|x-1\right|+\left|x-5\right|>8\)
Tìm x biếta,left(frac{4}{5}right)^{2x+7}text{}frac{625}{256}b,frac{7^{x+2}+7^{x+1}+7^x}{57}text{}frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}c,left(4x-3right)^4text{}left(4x-3right)^2d,frac{2x+3}{5x+2}text{}frac{4x+5}{10x+2}e,frac{3x-1}{40-5x}text{}frac{2x-3x}{5x-34}f,frac{15}{x-9}text{}frac{20}{y-12}text{}frac{40}{z-2x} và xytext{}1200
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Tìm x biết
a,\(\left(\frac{4}{5}\right)^{2x+7}\text{=}\frac{625}{256}\)
b,\(\frac{7^{x+2}+7^{x+1}+7^x}{57}\text{=}\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
c,\(\left(4x-3\right)^4\text{=}\left(4x-3\right)^2\)
d,\(\frac{2x+3}{5x+2}\text{=}\frac{4x+5}{10x+2}\)
e,\(\frac{3x-1}{40-5x}\text{=}\frac{2x-3x}{5x-34}\)
f,\(\frac{15}{x-9}\text{=}\frac{20}{y-12}\text{=}\frac{40}{z-2x}\) và \(xy\text{=}1200\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
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Tìm x, biết:
a) \(x\left(x-1\right)-x^2+2\text{x}=5\)
b) \(8\left(x-2\right)-2\left(3\text{x}-4\right)=2\)
c) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
d) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
tìm x, biết:
a) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
b) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
Tìm x:
a)\(\text{(x-5)(x+5)-(x+3)^2+3(x-2)^2=(x+1)^2-(x+4)(x-4)+3x^2}\)
\(\text{b)(2x+3)^2}+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
22 tháng 10 2021 lúc 19:52
Có thể dùng định lí Bezu nha
Tìm a,b sao cho
a) \(2x^3-x^2+ax+b\text{⋮}x^2-1\)
b) \(ax^3+bx^2+2x-1\text{⋮}x^2+5x-6\)
c) \(ax^{4\:}+bx^3+1\text{⋮}\left(x+1\right)^2\)
d) \(x^3-x-15\text{⋮}x^2+ax+b\)
e) \(x^3+ax+b\text{⋮}x^3+x-6\)
\(a,\Leftrightarrow2x^3-x^2+ax+b=\left(x-1\right)\left(x+1\right)\cdot a\left(x\right)\)
Thay \(x=1\Leftrightarrow2-1+a+b=0\Leftrightarrow a+b=-1\)
Thay \(x=-1\Leftrightarrow-2-1-a+b=0\Leftrightarrow b-a=3\)
Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\-a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=1\end{matrix}\right.\)
\(b,\Leftrightarrow ax^3+bx^2+2x-1=\left(x-1\right)\left(x+6\right)\cdot b\left(x\right)\)
Thay \(x=1\Leftrightarrow a+b+2-1=0\Leftrightarrow a+b=-1\)
Thay \(x=-6\Leftrightarrow-216a+36b+12-1=0\Leftrightarrow216a-36b=11\)
Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\216a-36b=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{25}{252}\\b=-\dfrac{227}{252}\end{matrix}\right.\)
\(c,\Leftrightarrow ax^4+bx^3+1=\left(x+1\right)^2\cdot c\left(x\right)\)
Thay \(x=-1\Leftrightarrow a-b+1=0\Leftrightarrow b=a+1\)
\(\Leftrightarrow ax^4+\left(a+1\right)x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^4+ax^3+x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^3\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(ax^3+x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow ax^3+x^2-x+1⋮\left(x+1\right)\)
Thay \(x=-1\Leftrightarrow-a+1+1+1=0\Leftrightarrow a=3\Leftrightarrow b=4\)
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Các bạn giúp mình bài toán sau
\(\left(x+2\right)^3\text{-}\left(x+1\right)\left(x^2\text{-}x+1\right)=10\)
\(\left(x\text{-}1\right)^3\text{-}\left(x\text{-}2\right)\left(x^2+x+4\right).3x\left(x+1\right)=0\)
\(\left(x\text{-}3\right)^2\text{-}\left(x\text{-}2\right)\left(x^2+2x+4\right)\text{-}9x\left(x\text{-}1\right)=0\)
đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)
\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)
TH1: \(x-1=0\Leftrightarrow x=1\)
TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)