Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

= (1 - 2x)(1 + 2x)

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)

\(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2}{1-x^2}\)

\(=\frac{1}{x+1}+\frac{1}{x-1}-\frac{2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x-1+x+1-2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{2x-2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{2}{x+1}\)

a: Ta có: \(\left(x-3\right)^2-x\left(x+5\right)=9\)

\(\Leftrightarrow x^2-6x+9-x^2-5x=9\)

\(\Leftrightarrow x=0\)

b: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)

mik lớp 6 bạn

x3 , y3 là gì vậy bn ???

Phần a dễ bạn tự làm nha!!! :))

b, Ta có: \(\Delta^'=\left[-\left(m+1\right)\right]^2-2m=m^2+2m+1-2m=m^2+1>0\forall m\)

=> PT luôn có 2 nghiệm phân biệt

Theo Vi-ét, ta có: \(\hept{\begin{cases}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{cases}}\)

Ta có: \(\sqrt{x_1}+\sqrt{x_2}=\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=2\)

\(\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=2\)

\(\Leftrightarrow x_1+x_2-2+2\sqrt{x_1x_2}=0\)

\(\Leftrightarrow2\left(m+1\right)-2+2\sqrt{2m}=0\)

\(\Leftrightarrow2m+2\sqrt{2m}=0\)

\(\Leftrightarrow m+\sqrt{2m}=0\)

\(\Leftrightarrow\sqrt{m}\left(\sqrt{m}+\sqrt{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{m}=0\\\sqrt{m}+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\\sqrt{m}=-\sqrt{2}\end{cases}}}\)

Vậy: m = 0

=.= hk tốt!!

a) Khi m=1 thì pt<=>x²-4x+2=0

Có:\(\Delta\)'=(-2)^2-2=2>0=>pt có 2 nghiệm là x₁=\(2+\sqrt{2}\)và x₂=2-\(\sqrt{2}\)

b)Để pt có nghiệm thì \(\Delta\)'=(m+1)²-2\(\ge\)0<=>m\(\ge\)\(\sqrt{2}\)-1

Theo định lý Viète thì:x₁+x₂=2(m+1)=\(\sqrt{2}\)<=>\(\frac{\sqrt{2}-2}{2}\)