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bài 1:
a. x2 - 5=0
=>x2 = 0+5 = 5
=> x = \(\sqrt{5}\)
vậy x= \(\sqrt{5}\)
sorry biết mỗi a thôi
a) x2 - 5 = 0
x2 = 0 + 5
x2 = 5
=> x = \(\sqrt{5}\)
Vậy ...
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
Vậy tập nghiệm của pt là \(S=\left\{4;5\right\}\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
Vậy tập nghiệm của pt là \(S=\left\{-6;7\right\}\)
Bài 1 :
a, \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)
\(=x^2+6x+9+x^2-6x+9+2x^2-18\)
\(=4x^2\)
b, \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)
\(=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9=8\)
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
Bài 1:
a) \(3x-6y=3\left(x-2y\right)\)
b) \(\frac{2}{5}x^2+x^3+x^2y\)
\(=x^2\left(\frac{2}{5}+x+y\right)\)
c) \(14x^2y-21xy^2+28x^2y^2\)
\(=xy\left(14x-21y+28xy\right)\)
d) \(\frac{2}{3}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)
\(=\left(y-1\right)\left(\frac{2}{3}x-\frac{2}{5}y\right)\)
e) \(10x\left(x-y\right)-8y\left(y-x\right)\)
\(=10x\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(x-y\right)\left(10x+8y\right)\)
f) \(4x^2-25\)
\(=\left(2x\right)^2-5^2\)
\(=\left(2x-5\right)\left(2x+5\right)\)
g) \(6x-9-x^2\)
\(=-x^2+6x-9\)
\(=-\left(x+3\right)^2\)
h) \(x^3+y^3+z^3-3xyz\)
Đề sai ?? ko tách đc nữa
Bài 2:
a) \(15.91,5+150.0,85\)
\(=15.\left(91,5+8,5\right)\)
\(=15.100\)
\(=1500\)
b) \(x\left(x-1\right)-y\left(1-x\right)\)
\(=x\left(x-1\right)+y\left(x-1\right)\)
\(=\left(x-1\right)\left(x+y\right)\)
\(=2000.4000\)
\(=8000000\)
c) \(87^2+73^2-27^2-13^2\)
\(=\left(87^2-27^2\right)+\left(73^2-13^2\right)\)
\(=\left(87-27\right)\left(87+27\right)+\left(73-13\right)\left(73+13\right)\)
\(=60.114+60.86\)
\(=60.\left(144+86\right)\)
\(=60.230\)
\(=82800\)
d) \(73^2-27^2\)
\(=\left(73-27\right)\left(73+27\right)\)
\(=46.100\)
\(=4600\)
e) f) tương tự
\(x^2-25x=0\)
\(\Rightarrow x\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
vậy_
\(\left(4x-1\right)^2-9=0\)
\(\Rightarrow\left(4x-1\right)^2-3^2=0\)
\(\Rightarrow\left(4x-1+3\right)\left(4x-1-3\right)=0\)
\(\Rightarrow\left(4x+2\right)\left(4x-4\right)=0\)
\(\Rightarrow2\cdot\left(2x+1\right)\cdot4\cdot\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}}\)
vậy_
Bài 2 :
a) \(3x^2-18x+27\)
\(=3\left(x^2-6x+9\right)\)
\(=3\left(x^2-2\cdot x\cdot3+3^2\right)\)
\(=3\left(x+3\right)^2\)
b) \(xy-y^2-x+y\)
\(=y\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(y-1\right)\)
c) \(x^2-5x-6\)
\(=x^2+x-6x-6\)
\(=x\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x-6\right)\)