tính:
x(x+4)(x+5)(x+9)+ 91
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ML
1
9 tháng 8 2017
A = 4 . 2 . 25 . 5 . 175
A = 22 . 2 . 52 . 5 . 52 . 7
A = 22+1 . 52+1+2 . 7
A = 23 . 55 .7
A = 175000
H
9 tháng 8 2018
\(B=4^2-10^4:\left(50\cdot273-50\cdot73\right)\)
\(B=4^2-10^4:\left[50\cdot\left(273-73\right)\right]\)
\(B=4^2-10^4:\left(50\cdot200\right)\)
\(B=4^2-10^4:10^4=4^2-1=15\)
\(C=3\times53\times6+2\times9\times87-18\times40\)
\(C=18\times53+18\times87-18\times40\)
\(C=18\times\left(53+87-40\right)\)
\(C=18\times100=1800\)
GL
2
TT
8 tháng 2 2021
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
8 tháng 2 2021
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
19 tháng 6 2023
1:
a: =>x:5=6484
=>x=32420
b: =>x+23839=91834
=>x=67995
6 tháng 12 2023
Bài 4:
a: xy=-2
=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)
=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)
b: \(\left(x-1\right)\left(y+2\right)=-3\)
=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)
=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)
Bài 3:
a: \(x\left(x+9\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b: \(\left(x-5\right)^2=9\)
=>\(\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)
c: \(\left(7-x\right)^2=-64\)
mà \(\left(7-x\right)^2>=0\forall x\)
nên \(x\in\varnothing\)
Bài 2:
a: \(\left(-31\right)\cdot x=-93\)
=>\(31\cdot x=93\)
=>\(x=\dfrac{93}{31}=3\)
b: \(\left(-4\right)\cdot x=-20\)
=>\(4\cdot x=20\)
=>\(x=\dfrac{20}{4}=5\)
c: \(5x+1=-4\)
=>\(5x=-4-1=-5\)
=>\(x=-\dfrac{5}{5}=-1\)
d: \(-12x+1=-4\)
=>\(-12x=-4-1=-5\)
=>\(12x=5\)
=>\(x=\dfrac{5}{12}\)
Theo mình thì thế này :
\(x\left(x+4\right)\left(x+5\right)\left(x+9\right)+91\)
= \(\left[x.\left(x+9\right)\right]\left[\left(x+4\right)\left(x+5\right)\right]+91\)
= \(\left(x^2+9x\right).\left(x^2+9x\right)+91\)
Đặt \(t=\left(x^2+9x\right)\)
=> \(t^2+91\)
=> \(\left(x^2+9x\right)^2+91\)
=> \(x^4+19x^2+81x^2+91\)
x(x+4)(x+5)(x+9)= 91
=> x. x( 4+5+9) =91
=> x.x . 18= 91
=> x.x= 73
=> \(x^2=\) 73
=> x=\(\sqrt{73}\)