Rút gọn biểu thức
a. B = \(\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
b. C = \(a:\left(b-2\right)-\left[\left(a^2+2a+1\right):\left(b^2-4\right)\right].\left[\left(b+2\right):\left(a+1\right)\right]\)
Rút gọn biểu thức
a. B = \(\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
b. C = \(a:\left(b-2\right)-\left[\left(a^2+2a+1\right):\left(b^2-4\right)\right].\left[\left(b+2\right):\left(a+1\right)\right]\)
\(A=\left(\dfrac{1}{2a-b}-\dfrac{a^2-1}{2a^3-b+2a-a^2b}\right)\div\left(\dfrac{4a+2b}{a^3b+ab}-\dfrac{2}{a}\right)\)
a) rút gọn biểu thức A
b)tính giá trị biểu thức A biết 4a^2+b^2=5ab a>b>0
Rút gọn biểu thức: \(B=\left(ab+bc+ca\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)
\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=2\left(a+b+c\right)\)
Bài 1 : Rút gọn biểu thức
a. A = \(\left(a-2\right):\left\{\dfrac{a^2-b^2}{a^3+b^3}.\left[a-\dfrac{a^2+b^2}{b}:\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\right]\right\}=\dfrac{a-2}{a}\)
b. B = \(1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\)
2. Chứng minh đẳng thức :
a. \(\left(\dfrac{6a+1}{a^2-6a}+\dfrac{6a-1}{a^2+6a}\right).\dfrac{a^2-36}{a^2+1}=\dfrac{12}{a}\)
b. \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
Bài 1 : Rút gọn biểu thức
a. A = \(\left(a-2\right):\left\{\dfrac{a^2-b^2}{a^3+b^3}.\left[a-\dfrac{a^2+b^2}{b}:\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\right]\right\}=\dfrac{a-2}{a}\)
b. B = \(1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\)
2. Chứng minh đẳng thức :
a. \(\left(\dfrac{6a+1}{a^2-6a}+\dfrac{6a-1}{a^2+6a}\right).\dfrac{a^2-36}{a^2+1}=\dfrac{12}{a}\)
b. \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
3. Chứng minh biểu thức không phụ thuộc vào biến :
a. A = \(\left(\dfrac{x}{x-y}-\dfrac{y}{x+y}\right):\left(\dfrac{x+y}{x-y}-\dfrac{2xy}{x^2-y^2}\right)\)
b. \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
Bài 1 : Rút gọn biểu thức
a. A = \(\left(a-2\right):\left\{\dfrac{a^2-b^2}{a^3+b^3}.\left[a-\dfrac{a^2+b^2}{b}:\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\right]\right\}=\dfrac{a-2}{a}\)
b. B = \(1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\)
2. Chứng minh đẳng thức :
a. \(\left(\dfrac{6a+1}{a^2-6a}+\dfrac{6a-1}{a^2+6a}\right).\dfrac{a^2-36}{a^2+1}=\dfrac{12}{a}\)
b. \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
3. Chứng minh biểu thức không phụ thuộc vào biến :
a. A = \(\left(\dfrac{x}{x-y}-\dfrac{y}{x+y}\right):\left(\dfrac{x+y}{x-y}-\dfrac{2xy}{x^2-y^2}\right)\)
b. \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
Bài 3:
\(a,A=\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)\left(x+y\right)}:\dfrac{x^2+2xy+y^2-2xy}{\left(x-y\right)\left(x+y\right)}\\ A=\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x+y\right)}{x^2+y^2}=1\\ b,=\left[\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\\ =\left(a+2\sqrt{a}+1\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\\ =\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)
3.P=\(\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)\):\(\left(\dfrac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
a)Rút gọn P
b)Tìm những giá trị nguyên của a để P có giá trị nguyên
Rút gọn biểu thức
a) \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\left(\sqrt{a+\sqrt{b}}\right)^2-4\sqrt{ab}}.\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) \(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)\(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
HELP ME PLSSSSSSSSSS
câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )
a
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)
