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( 2 3/4 - 1 4/5 ) x = 1
=> ( 11/4 - 9/5 ) x = 1
=> 19/20 x = 1
=> x = 1 : 19/20
=> x = 20/19
x^2 - 9 .( 3 - 5x ) = 0
=> x^-7 . ( 3 - 5x ) = 0
=> x^-7 = 0 hoặc 3 - 5x = 0
=> Mà x^-7 = 0 => x loại
=> 3 - 5x = 0
=> 5x = 3 - 0 = 3
=> x = 3 : 5
=> x = 3/5
a)12(x-1)=0
=>x-1=0
=>x=1
b) 3/5x-3/4=(-3)2-|-11|
3/5x-3/4=9-11=-2
3/5x=-2+3/4=-5/4
=>x=-5/4:3/5=-25/12
c)4(x-5/8)-3/4=0,25
4(x-5/8)=0,25+3/4=1
=>x-5/8=1/4
x=1/4+5/8=7/8
2) Số đó là
27:3/5=45
\(a,\frac{1}{2}-\frac{2}{5}+3x=\frac{3}{4}\)
\(\frac{1}{10}+3x=\frac{3}{4}\)
\(3x=\frac{3}{4}-\frac{1}{10}\)
\(3x=\frac{13}{20}\)
\(x=\frac{13}{60}\)
\(b,2\cdot\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(-10x+3x=\frac{4}{5}-\frac{3}{2}\)
\(-7x=\frac{-7}{10}\)
\(x=\frac{1}{10}\)
a)100-7.[x-5]=58
7.[x-5]=100-58=42
x-5=42:7=6
x=6+5
x=11
b)12.[x-1]:3=43+23
12 [x-1]:3=64+8
12.[x-1]:3=72
[x-1]:3=72:12=6
[x-1]=6.3=18
x=18-1
x=17
tao koooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo biếtttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt
a)x+15=-7
x=(-7)-15
x=-22
b)\(\left(5x-2\right).8^3=8^4\)
(5x-2).512=4096
(5x-2)=4096:512
(5x-2)=8
5x=8+2
5x=10
x=10:5
x=2
c)|x-1|=2
=>x-1=2 hoac x-1=-2
x=2+1 x=-2+1
x=3 x=-1
a/ => x = (-7) - 15 = -22
Vậy x = -22
b/ (5x - 2) . 83 = 84
=> 5x - 2 = 8
=> 5x = 10
=> x = 2
Vậy x = 2
c/ => x - 1 = 2 => x = 3
hoặc x - 1 = -2 => x = -1
Vậy x = 3 ; x = -1
\(a)x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
\(b)x^3-5x^2+8x-4\)
\(=x^3-x^2+x^2-5x^2+8x-4\)
\(=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
\(c)x^2-5x-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
Bài 3
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+y+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
=> 1/(x+y+z) = 2
<=> x + y + z = 1/2 <=> y + z = 1/2 - x (1)
.(y+z+1)/x = 2 <=> y + z + 1 = 2x
kết hợp với (1) => 1/2 - x + 1 = 2x
<=> x = 1/2 => y + z = 0 <=> y = -z
có (x+y-3)/z = 2
<=> x + y - 3 = 2z
<=> y - 2z = 5/2
do y = -z => -3z = 5/2 <=> z = -5/6
y = 5/6
Vậy nghiệm tìm được (x;y;z) = (1/2;5/6;-5/6)