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Vế trái lớn hơn hoặc bằng 0 nên 11x lớn hơn hoặc bằng 0.
\(\Rightarrow x\ge0\)
Do vậy chỉ cần bỏ dấu giá trị tuyệt đối là tính được.
Kết quả cuối cùng được \(x=\frac{10}{11}\)
a)
\(\left|x\right|-2\left|x\right|+3\left|x\right|=16+6\left|x\right|-19\)
\(\left|x\right|-2\left|x\right|+3\left|x\right|-6\left|x\right|=16-19\)
\(\left|x\right|.\left(1-2+3-6\right)=-3\)
\(\left|x\right|.\left(-4\right)=-3\)
\(\left|x\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
b,
2.(|x| - 5) - 15 = 9
\(2.\left(\left|x\right|-5\right)=9+15\)
\(2.\left(\left|x\right|-5\right)=24\)
\(\left|x\right|-5=24:2\)
\(\left|x\right|-5=12\)
\(\left|x\right|=12+5\)
\(\left|x\right|=17\)
\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
c,
|8 - 2x| + |4y - 16| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|8-2x\right|=0\\\left|4y-16\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}8-2x=0\\4y-16=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=8\\4y=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
d,
|x - 14| + |2y - x| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|x-14\right|=0\\\left|2y-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-14=0\\2y-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
2.Tìm x, y, z biết
a,
2.|3x| + |y + 3| + |z - y| = 0
\(\Rightarrow\left\{{}\begin{matrix}2.\left|3x\right|=0\\\left|y+3\right|=0\\\left|z-y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x\right|=0\\y+3=0\\z-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=0\\y=-3\\z=y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
b, (x - 3y)2 + | y + 4|= 0
\(\Rightarrow\left\{{}\begin{matrix}\left(x-3y\right)2=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-4\right)\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
b: =>3|x-5|=8+4=12
=>|x-5|=4
=>x-5=4 hoặc x-5=-4
=>x=9 hoặc x=1
d: =>2x+6=3-3x-2
=>2x+6=1-3x
=>5x=-5
hay x=-1
e: \(\Leftrightarrow x-3\inƯC\left(70;98\right)\)
\(\Leftrightarrow x-3\in\left\{1;2;7;14\right\}\)
mà x>8
nên \(x\in\left\{10;17\right\}\)
1. a, 3x + 2 \(⋮2x-1\)
Có 3(2x - 1) \(⋮2x-1\)
Và 2(3x - 2) \(⋮2x-1\)
=> 6x - 4 - 6x + 3 \(⋮2x-1\)
<=> -1 \(⋮2x-1\)
=> 2x - 1 \(\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> 2x = 2; 0
=> x = 1; 0 (thỏa mãn)
@Lớp 6B Đoàn Kết
1. b, x2 - 2x + 3 \(⋮x-1\)
<=> x(x - 2) + 3 \(⋮x-1\)
<=> x(x - 1) - x + 3 \(⋮x-1\)
<=> x(x - 1) - (x - 1) - 2 \(⋮x-1\)
<=> (x - 1)2 - 2 \(⋮x-1\)
<=> -2 \(⋮x-1\)
=> x - 1 \(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> x = 2; 0; 3; -1 (thỏa mãn)
@Lớp 6B Đoàn Kết
a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
d: =>x+5=0 và 3-y=0
=>x=-5 hoặc y=3
e: =>x-2=0 và y+1=0
=>x=2 và y=-1
Tim x biet:
(2x-1)5=(2x-1)7
Tim cac so nguyen to x , y thoa man:
(x-2)2.(y-3)2=4
trinh bay cach lam nhe
(2x-1)5=(2x-1)7
=>x=0;1;\(\frac{1}{2}\)
(x-2)2.(y-3)2=4
phân tích rồi kẻ bảng ra