Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

đề bài lỗi bn ơi

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

chịu

\(a,P=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\\ P=\sqrt{a}+2+2+\sqrt{a}=2\sqrt{a}+4\\ b,P=a+1\Leftrightarrow a+1=2\sqrt{a}+4\\ \Leftrightarrow a-2\sqrt{a}-3=0\\ \Leftrightarrow\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)=0\\ \Leftrightarrow\sqrt{a}=3\left(\sqrt{a}\ge0\right)\\ \Leftrightarrow a=9\left(tm\right)\)

\(M=a+\dfrac{4a+2ab+2b+b^2+4a-2ab-2b+b^2-4a}{\left(2-b\right)\left(2+b\right)}\\ M=a+\dfrac{4a+2b^2}{\left(2-b\right)\left(2+b\right)}=\dfrac{4a-ab^2+4a+2b^2}{\left(2-b\right)\left(2+b\right)}\\ M=\dfrac{8a-ab^2+2b^2}{4-b^2}\)

Ta có \(8a-b^2\left(a-2\right)=8a-\dfrac{a^2\left(a-2\right)}{\left(a+1\right)^2}=\dfrac{8a^3+16a^2+8a-a^3+2a^2}{\left(a+1\right)^2}=\dfrac{7a^3+18a^2+8a}{\left(a+1\right)^2}\)

\(4-b^2=4-\dfrac{a^2}{\left(a+1\right)^2}=\dfrac{4a^2+8a+4-a^2}{\left(a+1\right)^2}=\dfrac{3a^2+8a+4}{\left(a+1\right)^2}\)

\(\Leftrightarrow M=\dfrac{7a^3+18a^2+8a}{3a^2+8a+4}=\dfrac{a\left(7a+4\right)\left(a+2\right)}{\left(3a+2\right)\left(a+2\right)}=\dfrac{a\left(7a+4\right)}{3a+2}\)

a) Gợi ý: a 2 − 5 a + 4 = ( a − 1 ) ( a − 4 ) ; a 2 + 3 a − 4 = ( a − 1 ) ( a + 4 )  

Ta rút gọn được A = a + 1 a − 4  

b) Thay a = 5 vào biểu thức A tìm được A = 6

c) Ta biến đổi A = a + 1 a − 4 = 1 + 5 a − 4  

⇒ A ∈ ℤ ⇒ a ∈ − 1 ;   3 ;   5 ;   9

A xác định khi 5x-10 ≠0 <=> X ≠ 2b) A = x²-4x+4/5x-10= (x-2)²/5(x-2)= x-2/5c) x= -2018<=> A = -2018-2/5= -2020/5 = -404

Chúc bạn học tốt

a) ĐKXĐ: \(x\ne2\)

b) Ta có: \(A=\dfrac{x^2-4x+4}{5x-10}\)

\(=\dfrac{\left(x-2\right)^2}{5\left(x-2\right)}\)

\(=\dfrac{x-2}{5}\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)