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a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)
b) \(\left|x+2\right|+\left|y+5\right|=0\)
Ta có :
\(\left|x+2\right|\ge0\)
\(\left|y+5\right|\ge0\)
\(\Rightarrow\left|x+2\right|+\left|y+5\right|\ge0\)
Mà đề cho \(\left|x+2\right|+\left|y+5\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+2\right|=0\\\left|y+5\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+2=0\\y+5=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-2\\y=-5\end{cases}}}\)
(x + 2)(y - 3) = 5 = 1.5 = 5.1 = (-1).(-5) = (-5).(-1)
Xét 4 trường hợp , ta có :
\(\left(1\right)\hept{\begin{cases}x+2=1\\y-3=5\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=8\end{cases}}}\)
\(\left(2\right)\hept{\begin{cases}x+2=5\\y-3=1\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=4\end{cases}}\)
\(\left(3\right)\hept{\begin{cases}x+2=-1\\y-3=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=-2\end{cases}}\)
\(\left(4\right)\hept{\begin{cases}x+2=-5\\y-3=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-7\\y=2\end{cases}}\)
a) \(\left(x+2\right)\left(y-3\right)=5\)
Ta có bảng sau:
| x + 2 | 1 | 5 | -1 | -5 |
| y - 3 | 5 | 1 | -5 | -1 |
| x | -1 | 3 | -3 | -7 |
| y | 8 | 4 | -2 | 2 |
Vậy cặp số \(\left(x;y\right)\) là \(\left(-1;8\right);\left(3;4\right);\left(-3;-2\right);\left(-7;2\right)\)
b) \(\left|x+2\right|+\left|y+5\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+2\right|=0\\\left|y+5\right|=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+2=0\\y+5=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-2\\y=-5\end{matrix}\right.\)
Vậy \(x=-2;y=-5\)
c) tương tự b
d) sai đề
a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)
b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
c, \(\left(x-3\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d, \(\left(x-3\right)x-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)
\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
g, \(x^2+6x-7=0\)
\(\Rightarrow x^2-x+7x-7=0\)
\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
h,\(2x^2+5x-7=0\)
\(\Rightarrow2x^2-2x+7x-7=0\)
\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)
b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy \(x=3;x=\dfrac{5}{2}\)
d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)
e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)
câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha
g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)
\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)
h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)
\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)
a) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
Vì \(x^2-5>x^2-25\) nên \(\left\{{}\begin{matrix}x^2-5>0\\x^2-25< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2>5\\x^2< 25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{5}< x< -\sqrt{5}\left(vl\right)\\-5< x< 5\end{matrix}\right.\)
b) \(\left(x+5\right)\left(9+x^2\right)< 0\)
Vì \(9+x^2>0\) nên \(x+5< 0\Leftrightarrow x< -5\)
c) \(\left(x+3\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\) nên \(x+3=0\Leftrightarrow x=-3\)
d) \(\left(x+5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-2\\x=2\end{matrix}\right.\)
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 \(\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
1
= 1