15x-13= 12x-(-8)
Giúp mk vs !!
a) x^2-15x-16
b) x^2-12x-13
a)
=x2 - x +16x -16
=x(x-1) +16(x-1)
=(x+16)(x-1)
b)
=x2- x + 13x - 13
=x(x-1) + 13(x-1)
=(x+13)(x-1)
Tính giá trị biểu thức:
a. x^3 + 6x² + 12x + 8 tại x = 98
b. B = x³ – 9x²+ 27x – 27 tại x = 13
c. C= x³ - 3x^2+ 3x + 3 tại x= 11
d. D= x³ + 15x² + 75x + 126 tại x = 15
giúp mình với thank
Tìm x:
-5 ( 3x - 7 ) - ( -15x +3 ) - ( 12-x )= -4
-3 ( 4x -2 ) - ( -12x +8 ) - (-x)=0
a)-5 ( 3x - 7 ) - ( -15x +3 ) - ( 12-x )= -4
=>-15x+35+15x-3-12+x=-4
=>(-15x+15x)+(35-3-12)+x=-4
=>0+20+x=-4
=>20+x=-4
=>x=-4-20
=>x=-24
b)-3 ( 4x -2 ) - ( -12x +8 ) - (-x)=0
=>-12x+6+12x-8+x=0
=>(-12x+12x)+(6-8)+x=0
=>0+(-2)+x=0
=>(-2)+x=0
=>x=0-(-2)
=>x=2
GIẢI PT: 1) -2x^4 + 8x^3 - 3x^2 - 4x +4 =0
2) -3x^4 + 12x^3 - 26x^2 + 28x +8 =0
3) -2x^4 +12x^3 - 15x^2 -9x -1 =0
4) 3x^4 - 5x^3 - 16x^2+ 15x + 27 =0
mk mới lớp 6 thôi ,lớp 9 mình .......mình.........chịu (I VERY SORRY YOU!!)
mình lớp 9 nhưng mình lười giải vì " QUÁ NHIỀU " lười viết
15x+3-4x=17x-12x=43
\(15x+3-4x=17x-12x+43\)
\(\Leftrightarrow15x-4x+3=x\left(17-12\right)+43\)
\(\Leftrightarrow x\left(15-4\right)+3=5x+43\)
\(\Leftrightarrow11x+3=5x+43\)
\(\Leftrightarrow6x=40\)
\(\Leftrightarrow x=40\div6\)
\(\Leftrightarrow x=\frac{40}{6}\)
\(\Leftrightarrow x=\frac{20}{3}\)
12x+15x = 135
Tìm x
12x+15x= 135
(12+15)x = 135
27x = 135
x= 135:27
x=5
12x + 15x = 135
12+15x = 135
27x=135
x=135:27
x=5
12x + 15x = 135
12 + 15x = 145
27x = 135
x = 135 : 27
x = 5
x^2-5x+6
x^2-7x+12
x^2+x-12
x^2-9x+20
2x^2-3x-2
4x^2-7x-2
4x^2+15x+9
\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)
\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)
\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)
\(2x^2-3x+2=2\left(x+\dfrac{1}{2}\right)\left(x-2\right)\)
\(4x^2-7x-2=4\left(x-2\right)\left(x+\dfrac{1}{4}\right)\)
\(4x^2+15x+9=4\left(x+\dfrac{3}{4}\right)\left(x+3\right)\)
1+13/15x(0,5)^2x3+(8/15-25/100):1+23/24
=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+\(\frac{17}{60}\):1+\(\frac{23}{24}\)
=\(\frac{7}{15}\)x3+\(\frac{17}{60}\)+\(\frac{23}{24}\)
=\(\frac{7}{5}\)+\(\frac{17}{60}\)+\(\frac{23}{24}\)
=\(\frac{168}{120}\)+\(\frac{34}{120}\)+\(\frac{115}{120}\)
=\(\frac{317}{120}\)
\(12x+15x^2y=3x.4+3x.5xy=3x\left(4+5xy\right)\)
\(\text{Vậy } 12x + 15x^2y=3x(4+5xy)\)