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a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
a: x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b: 2x(x+3)=0
=>x(x+3)=0
=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
c: \(\left(6-x\right)\left(x+10\right)=0\)
=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)
d: \(\left(5x+20\right)\left(x^2+1\right)=0\)
=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)
=>5x=-20
=>x=-4
A=2.(1+2+22)+24.(1+2+22)+...+258.(1+2+22)
=2.7+24.7+...+258.7 chia het cho 7
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a: Bạn ghi lại đề nha bạn
b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)
=>\(30x+60-6x+30-24x=100\)
=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)
=>0x=100-90=10(vô lý)
c: \(\left(x-7\right)\left(x+3\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
d: -1<2x-1<4
=>\(-1+1< 2x< 4+1\)
=>0<2x<5
=>0<x<2,5
mà x nguyên
nên \(x\in\left\{1;2\right\}\)
câu 1:
a) 500-(300)-190+(-210)
= 500-300-190-210
= 200 - 210 -190
=-10 - 190
=-200
b) (-3)3 .5+12.(-6)
= -27.5 -72
=-135 - 72
=-207
c) 15.(-19-4)-19.(15-4)
= 15.(-23) - 19.11
=-345 - 209
=-554
câu 2: tìm x thuộc Z
a) 3x-2=3
=> 3x=3/2
=> x=1/2
b) x chia hết cho 5 và -7<x<11
=> x thuộc {-5;0;5;10}
Câu 1:
a) Ta có: \(500-\left(300\right)-190+\left(-210\right)\)
\(=500-300-190-210\)
\(=\left(500-300\right)-\left(190+210\right)\)
\(=200-400=-200\)
b) Ta có: \(\left(-3\right)^3\cdot5+12\cdot\left(-6\right)\)
\(=\left(-3\right)^3\cdot5-3\cdot4\cdot3\cdot2\)
\(=-5\cdot3^3-3^2\cdot8\)
\(=3^2\cdot\left(-5\cdot3-8\right)\)
\(=9\cdot\left(-15-8\right)=9\cdot\left(-23\right)=-207\)
c) Ta có: \(15\cdot\left(-19-4\right)-19\cdot\left(15-4\right)\)
\(=-15\cdot19-15\cdot4-15\cdot19+19\cdot4\)
\(=-30\cdot19+4\cdot4\)
\(=-2\cdot\left(15\cdot19+2\cdot4\right)\)
\(=-2\cdot\left(285+8\right)=-586\)
a, x.(x+7) b, (x-12).(x-3)=0
Ta có : x=0(t/m) hoặc x+7=0 ta có: x-12=0 hoặc x-3=0
x=0-7 x=0+12 x= 0+3
x=-7(t/m) x=12(t/m) x=3(t/m)
d, |2.n+1|=0 2.n+1=0 2.n=0+1 2n=1 n=1:2=0,5(ko t/m)
e, |2x+1|-19=-7 2x+1-19=-7 2x+(-18)=-7 2x=-7-(-18) 2x=11 x=11:2 x=5,5(ko t/m)
g, 2x+7 chia hết cho x+1 2x+7chia hết cho x+1 suy ra 2x+7 chia hết cho 2.(x+1) suy ra 2x+1 chia hết cho 2x+2
x+1 chia hết cho x+1
(2x+7)-(2x+2)= 5 , 5 chia hết cho x+1 x thuộc Z x+1 thuộc ước cuar5 = {+-1; +-5}
ta có x+1 1 -1 5 -5
x 0 -2 4 -6 (t/m)
vậy x thuộc 0; -2; 4; -6
a; x(x-7)=0 <=>x=0 hoặc x-7=0 b;x+12 hay x-12
x-7=0 =>x=7 Nếu: x+12 thì: Nếu: x-12 thì:
Vậy : x=0;7 (x+12)(x-3)=0 <=> x+12=0 hoặc x-3=0 (x-12)(x-3)=0
* x+12=0=>x=-12 <=> x-12=0 hoặc x-3=0
* x-3=0=>x=3 * x-12=0=>x=12
* x-3=0=>3
c;(-x+5)(3-x)=0<=> -x+5=0 hoặc 3-x=0 d; /2n+1/=0<=>2n+1=0 e; /2x+1/-19=-7
*-x+5=0=>x=5 *2n+1=0=>2n=-1 /2x+1/=12
*3-x=0 => x=3 vì 2n chia hết cho 2 mà -1 ko chia hết cho2 => 2x+1= -12;12
nên:ko có giá trị n *2x+1=-12 =>2x=-13
ko có x t/m
*2x+1=12=> 2x=11
ko có x t/m
f; x+7 chia hết cho n+2 g; 2x+7 chia hết cho x+1
=> x+2+5 chia hết cho n+2 =>2(x+1)+5 chia hết cho x+1
x+2 chia hết cho n+2 nên để x+7 chia hết cho n+2 thì : x+1 chia hết cho x+1 =>2(x+1) chia hết cho x+1 nên để:
5chia hết cho n+2 =>n+2=1;5 2x+7 chia hết cho x+1 thì: 5 chia hết cho x+1 =>x+1=1;5
*n+2=1=> n=-1 * x+1=1 => x=0
* n+2=5 => n=3 * x+1=5 =>x=4
Chúc bạn học giỏi!