TÌM X,Y/x+13/14/+/y-8/27/=0
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Vì: \(\left|x+\frac{13}{14}\right|\ge0\forall x;\left|y-\frac{8}{27}\right|\ge0\forall y\)
mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{8}{27}\right|=0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{8}{27}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{8}{27}\end{cases}}\)
Vậy \(x=-\frac{3}{4};y=\frac{8}{27}\)
b,
ta có: x-12/3 + y+8/23 + z+190/27 luôn lớn hơn 0 nên không thể nhỏ hơn 0
Để: |x-12/3| + |y+8/23| + |z+190/27| > 0
=> (+) x-12/3 = 0
=> x= 12/3
(+) y+8/23 = 0
=> y = -8/23
(+) z+190/27 = 0
=> z = -190/27
Vậy x = 12/3; y = -8/23; z = -190/27
k giúp mình
làm ơn
\(1,x^2+2xy+x+2y\)
\(=\left(x^2+2xy\right)+\left(x+2y\right)\)
\(=x\left(x+2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+1\right)\)
\(2,x^2-10x+25\)
\(=x^2-2.x.5+5^2\)
\(=\left(x-5\right)^2\)
Đợi mk chút ,mk có việc bận ,tối mk làm tiếp nha bn
\(3,x^3+3x^2+3x+1\)
\(=\left(x^3+1\right)+\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+3x\right)\)
\(=\left(x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x+1\right)\left(x+1\right)^2\)
\(=\left(x+1\right)^3\)
\(4,x^3-8\)
\(=x^3-2^3\)
\(=\left(x-2\right)\left(x^2+2x+4\right)\)
\(5,x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
\(6,x^3-\dfrac{1}{8}\)
\(=x^3-\left(\dfrac{1}{2}\right)^3\)
\(=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
\(7,x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
\(8,4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
\(9,49x^2-9\)
\(=\left(7x\right)^2-3^2\)
\(=\left(7x-3\right)\left(7x+3\right)\)
a) (-12) . (x - 14) + 7 . (3 - x) = 12.15
=> -12x + 168 + 21 - 7x = 180
=> -19x = -9
=> x = \(\frac{9}{19}\)
b) (2x - 8) . ( y - 2) = 0
=> \(\left\{\begin{matrix}2x-8=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=4\\y=2\end{matrix}\right.\)
c) x^2 - 75 = (-50)
=> \(x^2\)=25
=> \(\left[\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
d) (2x - 5)^2 - 1 = 0
=> ( 2x - 5 ) ^2 = 1
=> \(\left[\begin{matrix}2x-5=1\\2x-5=-1\end{matrix}\right.\)
=>\(\left[\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e) (3y + 6)^3 + 27 = 0
=> \(\left(3y+6\right)^3=-27\)
=> 3y + 6 = -3
=> y = -3
1a) 25 .8 - 12.5 + 272 :17 - 8
= 200 - 60 + 16 - 8
= 140+16-8
= 156 - 8
= 148
b) 125: 25 +14 - 142 : 71
= 5 + 14 - 2
= 19 - 2 = 17
c) 13. 17 - 256: 16 + 14 : 7 - 1
= 221 - 16 + 2 - 1
= 205 + 2 - 1
= 207 - 1 = 206
d) 15 . 24 - 14 . 5 (145 : 5 - 27)
= 360 - 70 (29- 27)
= 360 - 70 . 2
= 360 - 140
= 220
2a) 37 + x = 55
x= 55 - 37
x = 18
b) x- 31 = 24
x = 24 + 31
x = 55
c) 21 (x-11) = 21
x - 11 = 21 : 21
x = 1 +11
x = 12
d ) 9. (x - 29)= 0
x - 29 = 0 :9
x = 29
e) (x - 1954) . 5 = 50
x - 1954 = 50 : 5
x = 1964
f ) 30 . 60 - x = 30
1800 - x = 30
x = 1800 - 30
x = 1770
Viết mỏi tay quá huhuh
\(2x\left(2y-14\right)-8\left(y-7\right)=0\)
=>\(2x\cdot2\cdot\left(y-7\right)-8\left(y-7\right)=0\)
=>\(4x\left(y-7\right)-8\left(y-7\right)=0\)
=>\(\left(4x-8\right)\left(y-7\right)=0\)
=>\(\left\{{}\begin{matrix}4x-8=0\\y-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=7\end{matrix}\right.\)
Vì \(\left|x+\frac{13}{14}\right|\ge0;\left|y-\frac{8}{27}\right|\ge0\)
mà \(\left|x+\frac{13}{14}\right|+\left|y-\frac{8}{27}\right|=0\)
nên \(\left|x+\frac{13}{14}\right|=0\Rightarrow x+\frac{13}{14}=0\Rightarrow x=-\frac{13}{14}\)
\(\left|x-\frac{8}{27}\right|=0\Rightarrow x-\frac{8}{27}=0\Rightarrow x=\frac{8}{27}\)