cách của mình bạn hiểu đúng k? **** đi

4a² + 4a + 1 = (2a)² + 2.2a.1 + 1² = ( 2a + 1 )²

\(x^2y^4+2xy^2+1=\left(xy^2\right)^2+2xy^2+1=\left(xy^2+1\right)^2\)

\(2xy^2+x^2y^4+1=2xy^2+\left(xy^2\right)^2+1^2=\left(xy^2+1\right)^2\)

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=8.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

.....

\(=\left(3^{64}-1\right)\left(3^{64}+1\right)\)

\(=3^{128}-1\)

\(\Rightarrow A=\frac{3^{128}-1}{2}\)

Mọi người giúp em thêm bài 5abc, 8c với ạ!

khó thế nhờ (^o^)

a) 7x+7y=7(x+y)

b) 2x²y-6xy²=2xy(x-3y)

c)3x(x-1)+7x²(x-1)=x(x-1)(3+7x)

d)3x(x-4)+5x²(4-x)=(x-4)(3x-5x²)

=x(x-4)(3-5x)

e)6x⁴-9x³=3x³(2x-3)

f)5y⁸-15y⁶=5y⁶(y²-3)

1. \(x^4+6x^3+11x^2+6x+1=0\)

\(\Leftrightarrow x^4+6x^3+9x^2+2x^2+6x+1=0\)

\(\Leftrightarrow\left(x^2+3x+1\right)^2=0\)

\(\Leftrightarrow x^2+3x+1=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{2}=\frac{\sqrt{5}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=-\frac{3+\sqrt{5}}{2}\end{cases}}\)

2. \(x^4+x^3-4x^2+x+1=0\)

\(\Leftrightarrow\left(x^4+2x^2+1\right)+2.\frac{x}{2}\left(x^2+1\right)+\left(\frac{x}{2}\right)^2-\left(\frac{5}{2}x\right)^2=0\)

\(\Leftrightarrow\left(x^2+1+\frac{x}{2}\right)^2-\left(\frac{5}{2}x\right)^2=0\)

\(\Leftrightarrow\left(x^2-1\right)^2\left(x^2+3x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\x^2+3x+1=0\end{cases}}\)

+) ( x - 1 )² = 0

<=> x - 1 = 0

<=> x = 1

+) x² + 3x + 1 = 0

<=> ( x + 3/2 )² - 5/4 = 0

<=> ( x + 3/2 )² = 5/4

<=> \(\hept{\begin{cases}x+\frac{3}{2}=\frac{\sqrt{5}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=-\frac{3+\sqrt{5}}{2}\end{cases}}\)

Vậy pt có tập nghiệm \(S=\left\{1;\frac{-3+\sqrt{5}}{2};-\frac{3+\sqrt{5}}{2}\right\}\)

1a/ z² - 6z + 5 - t² - 4t = z² - 2 . 3z + 3² - 4 - t² - 4t = (z² - 2 . 3z + 3²) - (2² + 2 . 2t + t²) = (z - 3)² - (2 + t)²

b/ x² - 2xy + 2y² + 2y² + 1 = x² - 2xy + y² + y² + 2y + 1 = (x² - 2xy + y²) + (y² + 2y + 1) = (x - y)² + (y + 1)²

c/ 4x² - 12x - y² + 2y + 8 = (2x)² - 12x - y² + 2y + 3² - 1 = [ (2x)² - 2 . 3 . 2x + 3² ] - (y² - 2y + 1) = (2x - 3)² - (y - 1)²

2a/ (x + y + 4)(x + y - 4) = x² + xy - 4x + xy + y² - 4y + 4x + 4y + 16 = x² + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y² + 16

= x² + 2xy + y² + 4² = (x + y)² + 4²

b/ (x - y + 6)(x + y - 6) = x² + xy - 6x - xy - y² + 6y + 6x + 6y - 36 = x² + (xy - xy) + (-6x + 6x) + (6y + 6y) - y² - 36

= x² - y² + 12y - 6² = x² - (y² - 12y + 6²) = x² - (y² - 2 . 6y + 6²) = x² - (y - 6)²

c/ (y + 2z - 3)(y - 2z - 3) = y² -2yz - 3y + 2yz - 4z² - 6z - 3y + 6z + 9 = y² + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z² + 9

= y² - 6y - 4z² + 9 = (y² - 6y + 9) - 4z² = (y - 3)² - (2z)²

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x² + 4y² + 6yz - 2xy + 6yz + 9z² - 3xz = (2xy - 2xy) + (3xz - 3xz) - x² + (6yz + 6yz) + 9z² + 4y²

= -x² + 4y² + 12yz + 9z² = (4y² + 12yz + 9z²) - x² = [ (2y)² + 2 . 2 . 3yz + (3z)² ] - x² = (2y + 3z)² - x²