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\(x^2-10x+16\)
\(=\left(x^2-2x\right)-\left(8x-16\right)\)
\(=x.\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x-8\right)\)
Tham khảo nhé~
a, 8x2+10x =2x.(4x+5)
b, 4x2-8x+4 =4.(x2 -2x+1)=4.(x-1)2
c, 3x2 -3xy -5x +5y =(3x2-5x) - (3xy-5y) = x.(3x-5)- y.(3x-5)= (x-y).(3x-5)
d, x2+ 4x- 45=x2+ 9x- 5x- 45= x.(x+9)- 5.(x+9)=(x-5).(x+9)
a , 8 x 2 + 10 x
= 2 x ( 4 x + 5 )
b , 4 x 2 - 8 x + 4
= ( 2x ) 2 - 2 . 2 x . 2 + 2 2
= ( 2x + 2 ) 2
c ) 3 x 2 - 3 x y - 5 x + 5 y
= 3 x ( x - y ) - 5 ( x - y )
= ( 3x - 5 ) ( x - y )
d ) x 2 + 4x - 45
= x 2 + 2 x . 2 + 4 - 49
= ( x + 2 ) 2 - 49
= ( x + 2 ) 2 - 7 2
= ( x + 2 - 7 ) ( x + 2 + 7)
= ( x - 5 ) ( x + 9 )
a) \(x^2+6x+9\)
\(=\left(x+3\right)^2\)
\(=\left(x+3\right)\left(x+3\right)\)
b) \(10x-25-x^2\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
\(=-\left(x-5\right)\left(x-5\right)\)
c) \(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(\frac{1}{25}x^2-64y^2\)
\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)
\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)
b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)
c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)
d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)
c) Đặt \(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1,5=a\)
\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)
\(\Rightarrow A=a^2-0,25-6\)
\(\Rightarrow A=a^2-\frac{25}{4}\)
\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)
Thay \(a=x^2+3x+0,5\)vào A ta có :
\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)
\(A=\left(x^2+3x-2\right)\left(x^2+3x+3\right)\)
c, Đặt \(x^2+3x+2=a\)
Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)
\(=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a+2\right)\left(a-3\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Câu d làm tương tự .
Gợi ý : (x+3)(x+5) = x2 + 8x + 15
đặt bằng a rồi giải tiếp
Bài làm:
a) \(x^6-6x^4+12x^2-8\)
\(=\left(x^2-2\right)^3\)
b) \(x^2+16-8x=\left(x-4\right)^2\)
c) \(10x-x^2-25=-\left(x-5\right)^2\)
d) \(9\left(a-b\right)^2-4\left(x-y\right)^2\)
\(=\left[3\left(a-b\right)\right]^2-\left[2\left(x-y\right)\right]^2\)
\(=\left(3a-3b-2x+2y\right)\left(3a-3b+2x-2y\right)\)
e) \(\left(x+y\right)^2-2xy+1\)
\(=x^2+2xy+y^2-2xy+1\)
\(=x^2+y^2+1\)
sai sai
a. \(x^6-6x^4+12x^2-8=\left(x^2\right)^3-3\left(x^2\right)^2.2+3x^22-2^3=\left(x^2-2\right)^3\)
b. \(x^2+16-8x=x^2-8x+4^2=\left(x-4\right)^2\)
c. \(10x-x^2-25=10x-x^2-5^2=-\left(x-5\right)^2\)
d. \(9\left(a-b\right)^2-4\left(x-y\right)^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
e. \(\left(x+y\right)^2-2xy+1=x^2+2xy+y^2-2xy+1=x\left(x+2y\right)-y\left(y+2x\right)+2y^2+1\)
\(=x\left(x+y\right)-y\left(y+x\right)+xy-yx+2y^2+x=\left(x-y\right)\left(x+y\right)+2y^2+x\)
\(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5 \left(x-3\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
\(x^2-8x+15\)
\(=\left(x^2-3x\right)-\left(5x-15\right)\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
Tham khảo nhé~
a) \(x^2+8x+15=x^2+3x+5x+15=\left(x+3\right)\left(x+5\right)\)
b) \(x^2+3x+2=x^2+2x+x+2=\left(x+1\right)\left(x+2\right)\)
c) \(-x^2+7x-6=-x^2+x+6x-6=\left(-x+6\right)\left(x-1\right)\)
d) \(5x^3y-10x^2y^2+5xy^3=5xy\left(x^2-2xy+y^2\right)=5xy\left(x-y\right)^2\)
e) \(2x^2+7x-15=2x^2-3x+10x-15=\left(2x-3\right)\left(x+5\right)\)
\(x^4-4x^3+8x^2-16x+16 \)
\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
a,\(x^3-3x^2+3x-1-y^3=\left(x^3-1\right)-\left(3x^2-3x\right)-y^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)-y^3\)
\(=\left(x-1\right)\left(x^2-2x+1\right)-y^3\)
\(=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
....
\(x^2-10x+16\)
\(=x^2-2x-8x+16\)
\(=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)
\(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)=\left(x-3\right)\left(x-5\right)\)
c) x2 -10x + 16
= x2 - 2x - 8x + 16
= x.(x-2) - 8.(x-2)
= (x-2).(x-8)
d) x2 - 8x + 15
= x2 - 3x - 5x + 15
= x.(x-3) - 5.(x-3)
= (x-3).(x-5)