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\(D=50^2-49.51\)
\(\Leftrightarrow D=50^2-\left(50-1\right)\left(50+1\right)\)
\(\Leftrightarrow D=50^2-50^2+1=1\)
\(C=39^2+78.61+61^2\)
\(\Leftrightarrow C=39^2+2.39.61+61^2\)
\(\Leftrightarrow C=\left(39+61\right)^2=100^2=10000\)
a) 2x2-6x-x+3 = 2x(x-3) - (x-3) = (x-3)(2x-1)
b) x2-x-5x+5 = x(x-1) - 5(x-1) = (x-1)(x-5)
c) 5x(x-2y) + 2( x-2y)2 = (x-2y)(5x+2x-2y) = (x-2y)(7x-2y)
chú ý : (A-B)2=(B-A)2
d) 7x(4-y)2 - (4-y)3 = ( 16-8y+y2) (7x-4+y)
a) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(2x-1\right)\)
b) \(x^2-6x+5=x^2-5x-x+5=x\left(x-5\right)-\left(x-5\right)=\left(x-5\right)\left(x-1\right)\)
c)\(5x\left(x-2y\right)+2\left(2y-x\right)^2=5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
d) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)+\left(y-4\right)^3=\left(y-4\right)\left(7x-y-4\right)\)
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)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1:
\(f\left(x\right)=6x^2-x+1=0\)
\(\Leftrightarrow x\left(6x-1\right)=-1\)
\(\Leftrightarrow\) Khi x=1 thì 6x-1=-1 <=> 6x=0<=> x=0(không thõa mãn)
Khi x=-1 thì 6x-1=1 <=> 6x=2 <=> 2/6=1/3(không thõa mãn)
vậy phương trình đã cho vô ngiệm
Bài 2: Mk ko bt làm xin lỗi bạn
a) \(3x^2-2x\left(5+1,5x\right)+10x\)
\(=3x^2-10x-3x^2+10x=0\)
b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3,5x\right)\)
\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)
\(=-7x^2+7x\)
a)5x.(x-2y)+2.(2y-x)2 = 5x.(x-2y)+2.(x-2y)2 = (x-2y)[ 5x+2(x-2y)] = (x-2y)( 5x + 2x - 4y) = (x-2y)(7x-4y)
b) tương tự như trước, do là (A-B)^2 = a^2 - 2ab + b^2 = b^2 - 2ab + a^2 = ( b-a)^2
c)100x2-(x2+25)2= (10x)^2 -(x2+25)2= (10x + x^2 + 25)( 10x-x^2-25) = (x^2+10x+25)[-(x^2-10x+25)] =-1(x+5)^2 (x-5)^2
d)x2-xz-9y2+3y2= ( từ để suy nghĩ đã -_-)
e)x3-x2.5x+125 =
Bài 12:
a) \(\left(\dfrac{1}{2}x+4\right)^2\)
\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)
\(=\dfrac{1}{4}x^2+4x+16\)
b) \(\left(7x-5y\right)^2\)
\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)
\(=49x^2-70xy+25y^2\)
c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)
\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)
\(=y^4-36x^4\)
d) \(\left(x+2y\right)^2\)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
e) \(\left(x-3y\right)\left(x+3y\right)\)
\(=x^2-\left(3y\right)^2\)
\(=x^2-9y^2\)
f) \(\left(5-x\right)^2\)
\(=5^2-2\cdot5\cdot x+x^2\)
\(=25-10x+x^2\)
\(11,\)
\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=\left(7x+4\right).8=56x+32\)
\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-6xy\right)\)