Bài 1Thực hiện phép tính
a. \(\frac{1}{2}x\left(2-x\right)\)
b. \(\frac{x-5}{5-x}\)
Bài 2
a. Phân tích đa thức thành nhân tử : \(x+y-x^2+y^2\)
b. Tìm x biết \(x\left(x-3\right)+3x-1=0\)
Bài 3 Rút gọn
a. \(A=\frac{x\left(x+2\right)-x\left(x-2\right)+8}{x^2-4}:\frac{4}{x-2}\)
b. \(B=\left(1-\frac{a+b}{a-b}\right)\left(1-\frac{2b}{a+b}\right)\)
Bài 4 Rút gọn \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
Bài 1:
\(a, \dfrac{1}{2}x(2-x)=x-\dfrac{1}{2}x^2\)
\(b, \dfrac{x-5}{5-x}\)\(=-\dfrac{x-5}{x-5}\)\(=-1\)
Bài 2:
\(a, x+y-x^2+y^2=(x+y)-(x^2-y^2)=(x+y)-(x-y)(x+y)\)
\(=(x+y)(1-x+y)\)
\(b, x(x-3)+3x-1=0 \)
\(⇔x^2-3x+3x-1=0 \)
\(⇔x^2-1=0 \)
\(⇔(x-1)(x+1)=0 \)
\(⇔\left[\begin{array}{} x-1=0\\ x+1=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=1\\ x=-1 \end{array}\right.\)
Bài 3:
\(a,A=\dfrac{x(x+2)-x(x-2)+8}{x^2-4}:\dfrac{4}{x-2}\)
\(A=\dfrac{4x+8}{(x-2)(x+2)}.\dfrac{x-2}{4}\)
\(A=\dfrac{4(x+2)}{(x-2)(x+2)}.\dfrac{x-2}{4}\)
\(A=1\)
\(b, B=(1-\dfrac{a+b}{a-b})(1-\dfrac{2b}{a+b})\)
\(B=\dfrac{-2b}{a-b}.\dfrac{a-b}{a+b}\)
\(B=\dfrac{-2b}{a+b}\)
Bài 4:
\(C=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^{16}-1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^{32}-1)(2^{32}+1)=2^{64}-1\)
