Đại số lớp 8

\(\frac{4x^2-16}{x^2+2x}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x^2-4\right)}{x\left(x+2\right)}=\frac{A}{x}\)

\(\Leftrightarrow\frac{4\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x-2\right)}{x}=\frac{A}{x}\)

\(\Leftrightarrow4\left(x-2\right)=A\Leftrightarrow A=4x-8\)

a)Đk:\(2x^2+2x\ne0\Rightarrow2x\left(x+1\right)\ne0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x\ne0\\x\ne-1\end{array}\right.\) thì phân thức xác định

b)\(\frac{5x+5}{2x^2+2x}=\frac{5\left(x+1\right)}{2x\left(x+1\right)}=\frac{5}{2x}\). Giá trị phân thức =1

\(\Rightarrow\frac{5}{2x}=1\Rightarrow5=2x\Rightarrow x=\frac{5}{2}\)

kẻ phân số kiểu j đây bạn

a)Tam giác ABC cân tại A có AM là đường trung tuyến=> AM là đường cao => Góc AMC=90

Xét tứ giác AMCK có: AI=CI ( I là trung điểm AC); MI=IK(K đx M qua I)

=> Tứ giác AMCK là hình bình hành (d/h nhận biết hbh)

AMCK là hình bình hành có: góc AMC =90

=> AMCK là hình chữ nhật

b)AMCK là hình chữ nhật

=> AK=MC hay AK=BM; AK//MC hay AK//BM

Xét tứ giác AKMB có AK=BM;AK//BM

=> AKMB là hình bình hành

\(\left(\frac{x+2}{x+1}-\frac{2x}{x-1}\right).\frac{3x+3}{x}+\frac{4x^2+x+7}{x^2-x}\)

\(\Leftrightarrow\) \(\left(\frac{\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{3x+3}{x}+\frac{\left(x+1\right)\left(4x^2+x+7\right)}{\left(x+1\right)\left(x^2-x\right)}\)

\(\Leftrightarrow\) \(\left(\frac{\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{3x+3}{x}+\frac{\left(x+1\right)\left(4x^2+x+7\right)}{x.\left(x+1\right)\left(x-1\right)}\)

\(\Leftrightarrow\) \(\left(\frac{\left(3x+3\right)\left(x-1\right)\left(x+2\right)}{x.\left(x-1\right)\left(x+1\right)}\right)-\left(\frac{2x.\left(3x+3\right)\left(x+1\right)}{x.\left(x-1\right)\left(x+1\right)}\right)+\left(\frac{\left(x+1\right)\left(4x^2+x+7\right)}{x.\left(x-1\right)\left(x+1\right)}\right)\)

\(\Leftrightarrow\) (3x + 3)(x - 1)(x + 2) - 2x.(3x + 3)(x + 1) + (x + 1)(4x² + x + 7)

\(\Leftrightarrow\) 3x³ + 6x² − 3x − 6 − 6x³ − 12x² − 6x + 4x³ + 5x² + 8x + 7

\(\Leftrightarrow\) x³ − x² − x + 1

Vì bạn chỉ nói là tính thôi nên kết quả là x³ - x² - x + 1

Chúc bạn học tốt!

\(\frac{4x^2}{x^2+2x}=\frac{A}{x}\)\(\Rightarrow\frac{x\cdot4x}{x\left(x+2\right)}=\frac{A}{x}\)

\(\Rightarrow\frac{4x}{x+2}=\frac{A}{x}\Rightarrow4x^2=A\left(x+2\right)\)\(\Rightarrow A=\frac{4x^2}{x+2}\)

A=\(\frac{4x^2}{x+2}\)

đề đúng này hả /hoi-dap/question/139801.html nick kia giải r` kia

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

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

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

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

b) \(3x^2-7x-10=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

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

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

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

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

b) \(3x^2-7x-10=\left(3x^2+3x\right)-\left(10x+10\right)\)

= \(3x\left(x+1\right)-10\left(x+1\right)\)

= \(\left(x+1\right)\left(3x-10\right)\)

\(A=x^2+12x+36=x^2+12x+36+3=\left(x+6\right)^2+3\ge3\)

Dấu '=' xảy ra khi x=-6

\(B=9x^2-12x+4-4=\left(3x-2\right)^2-4\ge-4\)

Dấu '=' xảy ra khi x=2/3

\(C=-x^2+4x+1\)

\(=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

x³+5x²+3x-9

=x³-x²+6x²-6x+9x-9

=x²(x-1)+6x(x-1)+9(x-1)

=(x-1)(x²+6x+9)

=(x-1)(x+3)²

x^3+5x^2+3x+9 = x^3+6x^2-x^2+9x-6x-9

= x^2(x-1)+6x(x-1)+9(x-1)

= (x-1)(x^2+6x+9) = (x-1)(x+3)^2

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

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

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)