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a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)
Giờ chỉ cần thế x, y vô nữa là xong nhé.
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)
\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)
\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)
\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)
\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)
Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :
\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)
Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)
\(A=\frac{1}{x^2-x}+\frac{1}{x^2+x+1}+\frac{2x}{1-x^3}\)
\(A=\frac{1}{x.\left(x-1\right)}+\frac{1}{x^2+x+1}+\frac{2x}{\left(1-x\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-x}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1+x^2-x-2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x^3-1\right)}\)
Với x=10
\(\Rightarrow A=\frac{1}{10.\left(10^3-1\right)}\)
\(A=\frac{1}{10.999}\)
\(A=\frac{1}{9990}\)
Vậy \(A=\frac{1}{9990}\)tại x=10
\(ĐKXĐ:x\ne\pm1\)
a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{4x^2}{1-x^2}\right):\frac{2x^2-2}{x^2-2x+1}\)
\(\Leftrightarrow A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x^2}{x^2-1}\right):\frac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{4x-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{-4x\left(x-1\right)^3}{2\left(x-1\right)^2\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{-2x\left(x-1\right)}{\left(x+1\right)^2}\)
b) Thay x = -3 vào A, ta được :
\(A=\frac{\left(-2\right)\left(-3\right)\left(-3-1\right)}{\left(-3+1\right)^2}\)
\(\Leftrightarrow A=\frac{6.\left(-4\right)}{2^2}\)
\(\Leftrightarrow A=-6\)
c) Để A > -1
\(\Leftrightarrow-2x\left(x-1\right)>-\left(x+1\right)^2\)
\(\Leftrightarrow2x\left(x-1\right)< \left(x+1\right)^2\)
\(\Leftrightarrow2x^2-2x< x^2+2x+1\)
\(\Leftrightarrow x^2-4x-1< 0\)
\(\Leftrightarrow\left(x-2\right)^2-5< 0\)
\(\Leftrightarrow\left(x-2\right)^2< 5\)
Đoạn này bạn tự tìm giá trị x thỏa mãn là xong (Chú ý ĐKXĐ)
Bài 1:
a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)
\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)
b: Thay x=1/3 vào A, ta được:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
Bạn có thể giúp mình 2 câu còn lại dc kh ạ