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\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
a) 25x² - 10xy + y²
= (5x)² - 2.5x.y + y²
= (5x - y)²
b) 4/9 x² + 20/3 xy + + 25y²
= (2/3 x)² + 2.2/3 x.5y + (5y)²
= (2/3 x + 5y)²
c) 9x² - 12x + 4
= (3x)² - 2.3x.2 + 2²
= (3x - 2)²
d) Sửa đề: 16u²v⁴ - 8uv² + 1
= (4uv²)² - 2.4uv².1 + 1²
= (4uv² - 1)²
\(25x^2-10xy+y^2=\left(5x\right)^2-2.5x.y+y^2=\left(5x-y\right)^2\)
\(\dfrac{4}{9}x^2+\dfrac{20}{3}xy+25y^2=\left(\dfrac{2}{3}x\right)^2+2.\dfrac{2}{3}x.5y+\left(5y\right)^2=\left(\dfrac{2}{3}x+5y\right)^2\)
1, \(x^2+2xy+y^2=\left(x+y\right)^2\)
2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)
5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
1: =(x+y)^2
2: =(2x+3)^2
3: =(x+5/2)^2
4: =(4x-1)^2
5: =(x+1/2)^2
6: =(x-3/2)^2
7: =(x+1)^3
8: =(1/2x+1)^2
9: =(3y-1/3)^3
10: =(2x+y)^3
\(a,9a^2-6ab=1\) ( kiểm tra lại đề giúp mk)
\(b,25-10x+x^2\)
\(=5^2-2.5.x+x^2\)
\(=\left(5-x\right)^2=\left(x-5\right)^2\)
c, cx kiểm tra và viết rõ đề hộ mk ak
\(d,\left(x-y\right)^2-4\left(x-y\right)+4\)
\(=\left(x-y\right)^2-2.\left(x-y\right).2+2^2\)
\(=\left(x-y-2\right)^2\)
c/ \(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2\)
A)\(1-2x+x^2\)
\(=\left(1-x\right)^2\)
B)\(4y+4+y^2\)
\(=2^2+4y+y^2\)
\(=\left(2+y\right)^2\)
C)\(\frac{1}{16}+\frac{1}{2}x+x^2\)
\(=\left(\frac{1}{4}\right)^2+\frac{1}{2}x+x^2\)
\(=\left(\frac{1}{4}+x\right)\)
D)\(36x^2+12xy+y^2\)
\(=\left(6x+y\right)^2\)
a. Đề đúng phải là \(\frac{1}{4}a^2+2ab^2+4b^4\)hoặc \(\frac{1}{4}a^2+2ab+4b^2\)
Ở đây mình giải trường hợp 2, bạn dựa theo để giải trường hợp 1 nhé :))
\(\frac{1}{4}a^2+2ab+4b^2\)
\(=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a+2b\right)^2\)
b. \(25+10x+x^2\)
\(=x^2+2.x.5+5^2\)
\(=\left(x+5\right)^2\)
c. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)
\(=\left(y^4\right)^2-2.y^4.\frac{1}{3}+\left(\frac{1}{3}\right)^2\)
\(=\left(y^4-\frac{1}{3}\right)^2\)
a: \(\left(3x-1\right)\left(9x^2+3x+1\right)=27x^3-1\)
b: \(\left(1-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{5}+1\right)=1-\dfrac{x^3}{125}\)
c: \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
d: \(\left(4x+3y\right)\left(16x^2-12xy+9y^2\right)=64x^3+27y^3\)
