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a)x2(x+1)+2x(x+1)=0
=>(x2+2x)(x+1)=0
=>x(x+2)(x+1)=0
=>x=0 hoặc x+2=0 hoặc x+1=0
=>x=0 hoặc x=-2 hoặc x=-1
b)x(3x-2)-5(2-3x)=0
=>x(3x-2)+5(3x-2)=0
=>(x+5)(3x-2)
=>x+5=0 hoặc 3x-1=0
=>x=-5 hoặc \(x=\frac{2}{3}\)
c)\(\frac{4}{9}-25x^2=0\)
\(\Rightarrow\left(\frac{2}{3}\right)^2-\left(5x\right)^2=0\)
\(\Rightarrow\left(\frac{2}{3}-5x\right)\left(\frac{2}{3}+5x\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\frac{2}{3}-5x=0\\\frac{2}{3}+5x=0\end{array}\right.\)
\(\Rightarrow x=\pm\frac{2}{15}\)
d)\(x^2-x+\frac{1}{4}=0\)
\(\Rightarrow\frac{4x^2}{4}-\frac{4x}{4}+\frac{1}{4}=0\)
\(\Rightarrow\frac{4x^2-4x+1}{4}=0\)
\(\Rightarrow4x^2-4x+1=0\)
\(\Rightarrow\left(2x-1\right)^2=0\)
\(\Rightarrow x=\frac{1}{2}\)
a)17*91,5+170*0,85
=17*91,5+17*10*0,85
=17*91,5+17*8,5
=17*(91,5+8,5)
=17*100
=1700
b)20162-162
=(2016+16)(2016-16)
=2032*2000
=4064000
c)x(x-1)-y(1-x)
=x(x-1)+y(x-1)
=(x-1)(x+y)
Thay x=2001 và y=2999 đc:
=(2001-1)(2001+2999)
=2000*5000
=10 000 000
Trả lời tội ghê đó bạn nhưng mk gửi một bài mà sao bạn trả lời một câu vậy bạn nhưng dù sao vẫn cảm on nha
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)
\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2}\)
Bài1: Phân tích các đa thức sau thành nhân tử
a)36-4x2+4xy-y2
\(=6^2-\left(4x^2-4xy+y^2\right)\)
\(=6^2-\left(2x-y\right)^2\)
\(=\left(6+2x-y\right)\left(6-2x+y\right)\)
b)2x4+3x2-5
\(=2x^4-2x^2+5x^2-5\)
\(=2x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(2x^2+5\right)\left(x^2-1\right)\)
\(=\left(2x^2+5\right)\left(x-1\right)\left(x+1\right)\)
B1:a)\(36-4x^2+4xy-y^2=36-\left(4x^2-4xy+y^2\right)=6^2-\left(2x-y\right)^2\)
\(=\left(6-2x+y\right)\left(6+2x-y\right)\)
c)\(a^3-ab^2+a^2+b^2-2ab=a\left(a^2-b^2\right)+\left(a-b\right)^2\)\(=a\left(a-b\right)\left(a+b\right)+\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab+a-b\right)\)
d)\(x^2-\left(a^2+b^2\right)x+a^2b^2=x^2-a^2x-b^2x+a^2b^2\)\(=x\left(x-a^2\right)-b^2\left(x-a^2\right)=\left(x-a^2\right)\left(x-b^2\right)\)
e)\(x\left(x-y\right)+x^2-y^2=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)\(=\left(x-y\right)\left(x+x+y\right)=\left(x-y\right)\left(2x+y\right)\)
a) \(x^3+y^3+z^3-3xyz\)
\(=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Bài 1:
\(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left(\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right)\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)=2y\left(3x^2+y^2\right)\)
Bài 2:
\(\frac{4}{9}-25x^2=0\Leftrightarrow\left(\frac{2}{3}-5x\right)\left(\frac{2}{3}+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=\frac{2}{3}\\5x=-\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{15}\\x=-\frac{2}{15}\end{matrix}\right.\)
\(x^2-x+\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Bài 3:
\(A=17.91,5+17.8,5=17\left(91,5+8,5\right)=17.100=1700\)
\(B=\left(2016-16\right)\left(2016+16\right)=2000.2032=4064000\)
\(C=2001\left(2001-1\right)+2999\left(2001-1\right)\)
\(=2001.2000+2999.2000\)
\(=2000\left(2001+2999\right)\)
\(=2000.5000=10000000\)
Bài 1: Phân tích đa thức thành nhân tử:
a) (x + y)3 - (x - y)3
= ( x + y - x - y )[( x + y ) 2 - ( x + y )( x - y ) + ( x - y )2 ]
= 0 [( x + y ) 2 - ( x + y )( x - y ) + ( x - y )2 ]
= 0
Bài 2: Tìm x, biết:
a) \(\frac{4}{9}\) - 25x2 = 0
( \(\frac{2}{3}\))2 - ( 5x )2 = 0
( \(\frac{2}{3}\)+ 5x )( \(\frac{2}{3}\)- 5x ) = 0
\(\frac{2}{3}\)+ 5x = 0 ----> 5x = -\(\frac{2}{3}\) ---> x = -\(\frac{2}{15}\)
\(\frac{2}{3}\)- 5x = 0 --> 5x = \(\frac{2}{3}\) --> x = \(\frac{2}{15}\)
b) x2 - x + \(\frac{1}{4}\) = 0
x2 - 2. x . \(\frac{1}{2}\) + \(\left(\frac{1}{2}\right)\)2 = 0
( x - \(\frac{1}{2}\))2 = 0
x - \(\frac{1}{2}\) = 0
x = \(\frac{1}{2}\)
Bài 3: Tính nhanh giá trị các biểu thức sau:
a) 17.91,5 + 170.0,85
= 17.91,5 + 17.10.0,85
= 17.91,5 + 17.8,5
= 17 ( 91,5 + 8,5 )
= 170
b) 20162 - 162
= ( 2016 + 16 )( 2016 - 16 ).
= 2032.2000
= 4064000
c) x(x - 1) - y (1 - x) tại x = 2001 và y = 2999
x(x - 1) - y (1 - x)
= x(x - 1) + y ( x - 1 )
= ( x + y )( x - 1 )
Thay x = 2001 và y = 2999
( 2001 + 2999 )( 2001 - 1 )
= 5000. 2000
= 10000000