Tìm x, biết:
a) 5^x+5=25^x
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\(a,\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\3x-\dfrac{2}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
a: x=81-257=-176
b: x=-35+546=511
c: x=721+615=1336
d: x=-500:25=-20
e: x=-15*5=-75
f: x=612/9=68
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
a,\(Đkxđ:x\ge3\)
Ta có:
\(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow|x-3|=3-x\)
\(\Leftrightarrow x-3=\left[{}\begin{matrix}x-3\\3-x\end{matrix}\right.\)
\(TH1:x-3=x-3\Leftrightarrow0x=0\)
\(\Rightarrow\)\(x\in R\) và \(x\ge3\)
\(TH2:x-3=3-x\Leftrightarrow2x=6\Leftrightarrow x=3\)( ko thỏa mãn điều kiện)
vậy \(\left\{x\in R/x\ge3\right\}\)
b, \(Đkxđ:x\le\dfrac{5}{2}\)
Ta có:
\(\sqrt{25-20x+4x^2}+2x=5\)
\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}+2x=5\)
\(\Leftrightarrow\left|5-2x\right|=5-2x\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=5-2x\\5-2x=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in R\\x=\dfrac{5}{2}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(\left\{x\in R/x\le\dfrac{5}{2}\right\}\)
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
a: =>(x-5)(x+5)+(x-5)(3x-15)=0
=>(x-5)(x+5+3x-15)=0
=>(x-5)(4x-10)=0
=>x=5 hoặc x=5/2
c: =>x^3-3x^2+2x^2-6x-8x+24=0
=>(x-3)(x^2+2x-8)=0
=>(x-3)(x+4)(x-2)=0
=>\(x\in\left\{3;-4;2\right\}\)
a, bổ sung đề
\(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)
x=0.6378022262
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