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a)11/12 - (2/5 + x)= 2/3
2/5+x=11/12-2/3
2/5+x=1/4
x=1/4-2/5
x=-3/20
b) 2.x (x- 1/7)= 0
2x^2-2/7=0
2x^2=2/7
x^2=1/7
x=\(\sqrt{\frac{1}{7}}\) ;_\(\sqrt{\frac{1}{7}}\)
c)3/4+1/4:x=2/5
1/4:x=2/5-3/4=-7/20
x=1/4:-7/20=-5/7
d, (x- 1/2)2 =0
x-1/2=0
x=1/2
e, (2x -1)3= -8=(-2)^3
2x-1=-2
2x=-2+1=-1
x=-1/2
a) x2 + x = 0
=> x( x+ 1 ) = 0
=> x = 0
hoặc x = -1
b) b, (x-1)x+2 = (x-1)x+4
=> x + 2 = x + 4
=> 0x = 2 ( ktm)
Vậy ko có giá trị x nào thoả mãn đk
d) Ta có: x-1/x+5 = 6/7
=>(x-1).7 = (x+5).6
=>7x-7 = 6x+ 30
=> 7x-6x = 7+30
=> x = 37
Vậy x = 37
e, x2/ 6= 24/25
=> x2 . 25 = 6 . 24
⇒x2.25=144⇒x2.25=144
⇒x2=144÷25⇒x2=144÷25
⇒x2=5,76=2,42=(−2,42)⇒x2=5,76=2,42=(−2,42)
⇒x∈{2,4;−2,4}⇒x∈{2,4;−2,4}
Vậy x∈{2,4;−2,4}
Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
(5x+2)(x-7)=0
suy ra 5x+2=0 hoặc x-7=0
5x = -2
x = -2/5 hoặc x=7
\(x^2-x-6=0\Rightarrow x^2-2x+3x-6\\ \Rightarrow x\left(x-2\right)+3\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
hay x-2=0 hoặc x+3 = 0
vậy x = 2 hoặc x = -3
a)(2x-3)2=1<=> \(\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}< =>\orbr{\begin{cases}2x=4\\2x=2\end{cases}}}\)\(< =>\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
x=2 =>22.52=20y.5y <=>100 = 100y <=> y=1
x=1 => 2.5= 20y.5y <=>10=100y <=>y = 1/2
b)(4x-3)2+(y2-9)2\(\ge0\)
dấu = sảy ra khi \(\hept{\begin{cases}4x-3=0\\y^2-9=0\end{cases}< =>\hept{\begin{cases}4x=3\\y^2=9\end{cases}}}\)\(\hept{\begin{cases}x=\frac{3}{4}\\y=\pm3\end{cases}}\)
c) <=> (y-5)8 \(\le-\left(x+4\right)^7\) (1)
(y-5)8 >=0 với mọi y nên -(x+4)7 \(\ge\left(y-5\right)^8\ge0\)<=> (x+4)7\(\le0< =>x+4\le0< =>x\le-4\)
Khi đó (1) <=> y-5\(\le\sqrt[8]{-\left(x+4\right)^7}\) <=> y\(\hept{\begin{cases}y\le5-\sqrt[8]{-\left(x+4\right)^7}\\x\le-4\end{cases}}\)
a, x : (-1/2)^3 = -1/2
=> x : (-1/8) = -1/2
=> x = 4
vậy_
b, (3/4)^5.x = (3/4)^7
=> x = (3/4)^7 : (3/4)^5
=> x = (3/4)^2
=> x = 9/16
vậy-
c, (3/5)^8 : x = (-3/5)^6
=> (3/5)^8 : x = (3/5)^6
=> x = (3/5)^8 : (3/5)^6
=> x = (3/5)^2
=> x= 9 /25
a, \(\frac{22}{5}+\frac{1}{2}\cdot x^2=4\cdot\frac{8}{5}\)
=> \(\frac{22}{5}+\frac{1}{2}\cdot x^2=\frac{32}{5}\)
=> \(\frac{1}{2}\cdot x^2=\frac{32}{5}-\frac{22}{5}\)
=> \(\frac{1}{2}\cdot x^2=2\)
=> \(x^2=2:\frac{1}{2}=4\)
=> x = 2 hoặc x = -2
\(b,\frac{7}{2}-\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
=> \(\left|x-\frac{1}{3}\right|=\frac{7}{2}-\frac{5}{2}\)
=> \(\left|x-\frac{1}{3}\right|=1\)
=> \(x-\frac{1}{3}=1\)hoặc \(x-\frac{1}{3}=-1\)
=> x = 1 + 1/3 hoặc x = -1 + 1/3
=> x = 4/3 hoặc x = -2/3
c, \(\left[x-\frac{1}{2}\right]\left[-3-\frac{x}{2}\right]=0\)
=> x - 1/2 = 0 hoặc -3 - x/2 = 0
=> x = 0 + 1/2 hoặc x/2 = -3
=> x = 1/2 hoặc x = -6