Tìm x,biết:
\(8\times6+288:\left(x-3\right)^2=50\\ \)
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288:(x-3)2=2
=> (x-3)2=144
=>x-3=12(vì x thuộc N)
=> x=15
\(8.6+288:\left(x-3\right)^2=50\)
\(\Rightarrow48+288:\left(x-3\right)^2=50\)
\(\Rightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Rightarrow\left(x-3\right)^2=288:2=144\)
Mà \(\left(x-3\right)^2=144=12^2\)
\(\Rightarrow x-3=12\)
\(\Rightarrow x=12+3\)
\(\Rightarrow x=15\)
\(8\cdot6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=50-48=2\)
\(\left(x-3\right)^2=288:2=144\)
Vì x là số tự nhiên nên (x-3) là số tự nhiên.
\(\left(x-3\right)^2=12^2\)
\(x-3=12\)
\(x=12+3=15\)
Tìm \(x\);\(y\) biết rằng\(x^y+y^x-\left(x+y\right)^2-2^{\left(x+y\right)}=\left(x+y\right)\times6\)
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right):\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right):\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right):6069=\frac{1}{3}\)
\(\Rightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Rightarrow1488x-476,5x=20230\)
\(\Rightarrow1011,5x=20230\)
\(\Rightarrow x=20\)
Bài làm :
Ta có :
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right)\div\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right)\div\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right)\div6069=\frac{1}{3}\)
\(\Leftrightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Leftrightarrow1488x-476,5x=20230\)
\(\Leftrightarrow1011,5x=20230\)
\(\Leftrightarrow x=20\)
Vậy x=20
( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0
= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0
= ( 1 - 1/6 ) x 10 - x = 0
= 5/6 x 10 - x =0
= 25/3 - x =0
x = 25/3 - 0
x = 25/3
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)
\(\frac{5}{6}\times10-x=0\)
\(\frac{25}{3}-x=0\)
x =\(\frac{25}{3}-0=\frac{25}{3}\)
a) \(5\times\left(3+7\times x\right)=400\)
\(3+7\times x=80\)
\(7\times x=77\)
\(x=11\)
b) \(x\times37+x\times63=1200\)
\(x\times\left(37+63\right)=1200\)
\(x\times100=1200\)
\(x=12\)
c) \(x\times6+12:3=40\)
\(x\times6+4=40\)
\(x\times6=36\)
\(x=6\)
d) \(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=66\)
\(x+1=11\)
\(x=10\)
e) \(163:x+34:x=10\)
\(\left(163+34\right):x=10\)
\(197:x=10\)
\(x=19,7\)
Ta có : 8 . 6 + 288 : (x - 3)2 = 50
=> 48 + 288 : (x - 3)2 = 50
=> 288 : (x - 3)2 = 50 - 48
=> 288 : (x - 3)2 = 2
=> (x - 3)2 = 288 : 2
=> (x - 3)2 = 144
=> không tìm được x thỏa mãn điều kiện trên
8.6+288:(x-3)2 = 50
<=> 48+288:(x-3)2 = 50
<=> 288:(x-3)2 = 50-48 = 2
<=> (x-3)2 = 2.288 = 576 = 242
<=> \(\orbr{\begin{cases}x-3=24\\x-3=-24\end{cases}}\)
<=> \(\orbr{\begin{cases}x=24+3=27\\x=3-24=-21\end{cases}}\)
Vậy x thuộc { 27 ; -21 }
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
\(8.6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=50-48\)
\(288:\left(x-3\right)^2=2\)
\(\left(x-3\right)^2=288:2\)
\(\left(x-3\right)^2=144\)
\(\left(x-3\right)^2=12^2=\left(-12\right)^2\)
\(=>\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.=>\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Vậy x = 15 hoặc x = -9
cảm ơn