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đk 1 - x\(\ge\)0
=> x \(\le\)1
Khi đó |x - 2| = -(x - 2)
|x - 3| = -(x - 3)
....
|x - 9| = -(x - 9)
Khi đó |x - 2| + |x - 3| +... + |x - 9| = 1-x (8 cặp số ở VT)
<=> -(x - 2) + -(x - 3) + .... + -(x - 9) = 1 - x
=> -x + 2 - x + 3 - .... - x + 9 = 1 - x
=> -(x + x + ... x) + (2 + 3 + ... + 9) = 1 - x
8 hạng tử x 8 hạng tử
=> -8x + 44 = 1 - x
=> 7x = 43
=> x = 43/7
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minh nghi Qx phai la Q(x)= x^2+(2m+1)x+m^2.
thay x=1 vào P(x) thayx=-1 vào Q(x)
ta dc 1+2m+m^2=1-2m-1+m^2
<=>4m=-1
<=>m=-1/4
minh nghi Qx phai la Q(x)= x^2+(2m+1)x+m^2.
thay x=1 vào P(x) thayx=-1 vào Q(x)
ta dc 1+2m+m^2=1-2m-1+m^2
<=>4m=-1
<=>m=-1/4
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\(5^{x+2}+5^{x+1}+5^x=3785.\)
\(\Leftrightarrow5^x\cdot5^2+5^x\cdot5+5^x=3785.\)
\(\Leftrightarrow5^x\left(5^2+5+1\right)=3785\)
\(\Leftrightarrow5^x\left(25+5+1\right)=3785\)
\(\Leftrightarrow5^x31=3785\)
\(\Leftrightarrow5^x=\frac{3785}{31}=125\)
\(\Leftrightarrow5^x=5^3\)
\(\Leftrightarrow x=3\)
vậy x\(=3\)
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Bài 1: Tìm x, biết
a) (x + 12). (x - 3) = 0
\(\Leftrightarrow\left[\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
Vậy: \(x=-12;3\)
b) (- x + 5).(3 - x) = 0
\(\Leftrightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy: \(x=3;5\)
c)x.(2 + x).(7 - x) = 0
\(\Leftrightarrow\left[\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy: \(x=-2;0;7\)
Chúc bạn học tốt!!!
Bài 2:Tính nhanh
a) -183+127-117+103
\(=-183+103+127-117\)
\(=-80+10\)
\(=-70\)
b) 49-(187-51)+(-13+100)
\(=49-187+51-13+100\)
\(=49+51-187-13+100\)
\(=100-200+100\)
\(=100+100-200=0\)
c) 874.167-874.200+33.875
\(=874.\left(167-200\right)+33.875\)
\(=874.-33+33.875\)
\(=-33\left(874-875\right)\)
\(=-33.-1=33\)
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Có: \(x^2+5\left(2y-2020\right)^2=100=10^2+5\cdot0^2\)
\(\Rightarrow\hept{\begin{cases}x^2=10^2\\\left(2y-2020\right)^2=0^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\pm10\\y=1010\end{cases}}\)
Vậy \(\left(x;y\right)\in\left\{\left(10;1010\right);\left(-10;1010\right)\right\}\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=7450\)
\(\left(x+x+...+x\right)+\left(1+2+...+100\right)=7450\)
\(x\times100+\frac{\left(100+1\right)\times100}{2}=7450\)
\(x\times100+5050=7450\)
\(x\times100=7450-5050\)
\(x\times100=2400\)
\(x=2400:100\)
\(x=24\)
Học tốt #
(x+1)+(x+2)+...+(x+100)=7450
= (x.100)+(1+2+3+..+100)=7450
= (x.100)+5050 =7450
x.100=7450-5050
x.100=2400
x=2400:100
x=24