tìm x 2^3 x + 5^2 x = 2 ( 5^2 2^3 ) - 33
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\(2^3x+5^2x=2.\left(5^2+2^3\right)-33\)
\(8x+25x=2.\left(25+8\right)-33\)
\(8x+25x=2.33-33\)
\(8x+25x=66-33\)
\(8x+25x=33\)
\(x+\left(8+25\right)=33\)
\(x+33=33\)
\(x=0\)
\(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(15:\left(x+2\right)=\left(27+3\right):10\)
\(15:\left(x+2\right)=30:10\)
\(15:\left(x+2\right)=3\)
\(x+2=15:3\)
\(x+2=5\)
\(x=3\)
\(5^{10}:5^8+x=3^{20}:\left(-3\right)^{18}-2^{35}:\left(-2\right)^{33}\\ \Leftrightarrow5^2+x=3^2+\left(-2\right)^2\\ \Leftrightarrow25+x=13\\ \Leftrightarrow x=-12\)
x.3+1/2+3/2+5/2=33
x.3+9/2=33
x.3+4,5=33
x.3=33-4,5
x.3=28,5
x=28,5:3
x=9,5
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)
\(6-2\left(x-1\right)=4\)
\(\Rightarrow2\left(x-1\right)=6-4\)
\(\Rightarrow2\left(x-1\right)=2\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=1+1=2\)
________________
\(2\cdot\left(x-2\right)+1=7\)
\(\Rightarrow2\cdot\left(x-2\right)=7-1\)
\(\Rightarrow2\cdot\left(x-2\right)=6\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=3+2=5\)
_______________
\(\left(2\cdot x-3\right)+4=9\)
\(\Rightarrow2\cdot x-3=5\)
\(\Rightarrow2\cdot x=3+5\)
\(\Rightarrow2\cdot x=8\)
\(\Rightarrow x=\dfrac{8}{2}=4\)
________________
\(\left(3\cdot x-2\right)-1=3\)
\(\Rightarrow3\cdot x-2=3+1\)
\(\Rightarrow3\cdot x-2=4\)
\(\Rightarrow3\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{3}=2\)
a: =>2(x-1)=2
=>x-1=1
=>x=2
b: =>2(x-2)=6
=>x-2=3
=>x=5
c; =>2x-3=5
=>2x=8
=>x=4
d: =>3x-2=4
=>3x=6
=>x=2
e: =>2(6-x)=4
=>6-x=2
=>x=4
f: =>x-2=5
=>x=7
g: =>10-2x=4
=>2x=6
=>x=3
h: =>2x+4=3
=>2x=-1
=>x=-1/2
j: =>x+2=12
=>x=10
l: =>2x+3=3
=>2x=0
=>x=0
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
23.x+52.x=2.(52.23)-33
8.x+25.x=2.(25.8)-33
x.(25+8)=2.200-33
x.33=400-33
x.33=367
x=367:33
x=367/33