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a, Vì 1983 > 0
Mà 1983.(x-7) > 0
=> x-7 > 0
=> x > 7
5giá trị của x là : 8;9;10;11;12
b, Vì -2010 < 0
Mà (-2010).(x+3) > 0
=> x+3 < 0
=> x < -3
5 giá trị của x là : -4;-5;-6;-7;-8
Tk mk nha
mình sẽ cho cac ban dung 3 k nhung phai giai ro ra nhe tung buoc mot
`#040911`
`a)`
`6 \times x - 5 = 613`
`=> 6 \times x = 613 + 5`
`=> 6 \times x = 618`
`=> x = 618 \div 6`
`=> x = 103`
Vậy, `x = 103`
`b)`
`12 \times x + 3 \times x = 30`
`=> x \times (12 + 3) = 30`
`=> x \times 15 = 30`
`=> x = 30 \div 15`
`=> x = 2`
Vậy, `x = 2`
`c)`
`125 - 25 \times (x - 1) = 100`
`=> 25 \times (x - 1) = 125 - 100`
`=> 25 \times (x - 1) = 25`
`=> x - 1 = 25 \div 25`
`=> x - 1 = 1`
`=> x = 1 + 1`
`=> x = 2`
Vậy, `x = 2`
`d)`
`(x - 2) \times (9x - 4) = 0?`
`=>`
TH1: `x - 2 = 0`
`=> x = 0 + 2`
`=> x = 2`
TH2: `9x - 4 = 0`
`=> 9x = 4`
`=> x = 4/9`
Vậy, `x \in {2; 4/9}.`
\(a,6x-5=613\\ \Leftrightarrow6x=618\\ \Leftrightarrow x=103\\ b,12x+3x=30\\ \Leftrightarrow15x=30\\ \Leftrightarrow x=2\\ c,125-25\left(x-1\right)=100\\ \Leftrightarrow25\left(x-1\right)=25\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,\left(x-2\right)\cdot\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{9}\end{matrix}\right.\)
5(x+4)-3(x-2)=x
\(\Leftrightarrow5x+20-3x+6=x\)
\(\Leftrightarrow2x+26=x\)
\(\Leftrightarrow2x-x=-26\)
\(x=-26\)
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,\(\left(x-2\right)\left(x+2\right)=0\)
\(< =>\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)
b,\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)
\(< =>\orbr{\begin{cases}\frac{3x}{4}-\frac{2}{4}=0\\\frac{3x}{12}+\frac{16}{12}=0\end{cases}}\)
\(< =>\orbr{\begin{cases}3x-2=0\\3x+16=0\end{cases}}< =>\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)
\(\left(x-2\right)\left(x+2\right)=0\)
=> \(\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)
=> \(\orbr{\begin{cases}\frac{3}{4}x-\frac{1}{2}=0\\0,25x+\frac{4}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)
a) vì 81=34=(-3)4
=> x=3
hoặc x=-3
b) 2^x+3 +2^x=144
=> 2x.23+2x=144
=>2x.(23+1)=144
=>2x.9=144
=>2x=144:9=16
mà 16=24
=> x=4
c)x2-4x=0
=>x.(x-4)=0
=> x=0
hoặc x-4=0 => x=4
=> x=0
hoặc x=4
d)(5^4 + 4^7) (8^9-2^7) (2^4 - 4^2)
=(5^4 + 4^7) (8^9-2^7) .(16-16)
=(5^4 + 4^7) (8^9-2^7) .0
=0
e)4^3.51+4^3.49+2.(3^2 +4^2 )
=43.(51+49)+2.(9+16)
=64.100+2.25
=6400+50
=6450
f)3n+3+3n=252
3n.33+3n=252
3n.(33+1)=252
3n.28=252
=>3n=252:28=9
mà 9=32
=>x=2
a,x+7>-3
=>-3<x+7<3
=>-10<x<-4
=>x\(\in\){-9;-8;-7;-6;-5}
b,(x-1)(x+2)\(\le\)0
TH1:(x-1)(x+2)=0
=>x-1=0
hoặc x+2=0
=>x=1
hoặc x=-2
TH2:(x-1)(x+2)<0
=>(x-1)(x+2) trái dấu
K/n1:(x-1)>0;(x+2)<0
=>x>1 và x<-2(vô lí)
K/n2:(x-1)<0;(x+2)>0
=>x<1;x>-2(thỏa mãn)
=>-2<x<1
=>x\(\in\){1;-2;0;-1}
c,(x-3)(x+1)=0
=>x-3=0
hoặc x+1=0
=>x=3
hoặc x=-1
=>x\(\in\){3;-1}
x + 7 > - 3
Mà -10 + 7 = -3
=> x > - 10