Question

[Thân Thị Phương Trang]

f) 2a²b+4ab²-a²c+ac²-4b²c+2bc²-4abc

g) y(x-2z)²+8xyz+x(y-2z)²-2z(x+y)²

h) x⁵-5x³+4x

i) x³-11x²+30x

Bài 2: tìm x

a)5x(x-1)=x-1

b)2(x+5)-x²-5x=0

c)x³-1/4x=0

Bài 3: tính nhanh

1)x² +1/2x+1/16 tại x=49,75

2) x² -y²-2y-1 tại x=93 và y=6

Answer 1

[Nguyễn Huy Tú]

Bài 2:

a, \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)

Vậy...

b, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy...

c, \(x^3-\dfrac{1}{4}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy...

Bài 3:

1, Đặt \(A=x^2+\dfrac{1}{2}x+\dfrac{1}{16}=x^2+\dfrac{1}{4}.x.2+\dfrac{1}{16}\)

\(=\left(x+0,25\right)^2\)

Thay x = 49,75 vào A ta có:
\(A=50^2=2500\)

2, tương tự

Answer 2

[Thân Thị Phương Trang]

bài 1 bn ơi