2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
Cho biểu thức:\(P=\left(\dfrac{2\text{x}}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3\text{x}}\right)v\text{ới}x\ne\pm3;x\ne0;x\ne5\)
1, Chứng minh \(P=\dfrac{x}{x-5}\)
1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)
\(=\left(\dfrac{2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x\cdot\left(x-3\right)}\right)\)
\(=\dfrac{2x-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2\left(x-3\right)-x+1}{x\left(x-3\right)}\)
\(=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x-3\right)}{2x-6-x+1}\)
\(=\dfrac{x}{x-5}\)
Cho a,b,c và x,y,z khác nhau và khác 0
CMR: \(\text{Nếu }\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\text{Thì }\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
\(\dfrac{-7}{x}\text{=}\dfrac{-21}{x-34}\) \(\dfrac{4-x}{-5}\text{=}\dfrac{-5}{4-x}\)
\(\dfrac{3}{x+2}\text{=}\dfrac{5}{2x+1}\) \(\dfrac{1}{2}\text{=}\dfrac{x+1}{3x}\)
\(\dfrac{-3}{x+1}\text{=}\dfrac{4}{2-2x}\)
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)
4-x=5 hoặc 4-x=-5
x=-1 hoặc x=9
Cho :\(\dfrac{a}{b}=\dfrac{c}{d}CMR:\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}v\text{à}\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Đặt ; \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\) Ta có; \(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}\)
Rút gọn biểu thức C = \(\dfrac{\text{x}-\dfrac{\text{1}}{\text{x}^{\text{2}}}}{\text{1}+\dfrac{1}{x}+\dfrac{1}{x^2}}\)
Rút gọn biểu thức C = \(\dfrac{\text{x}-\dfrac{\text{1}}{\text{x}^{\text{2}}}}{\text{1}+\dfrac{1}{x}+\dfrac{1}{x^2}}\)
\(C=\dfrac{x-\dfrac{1}{x^2}}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}\)
Đk: \(x\ne0\)
\(\Rightarrow C=\dfrac{\dfrac{x^3-1}{x^2}}{\dfrac{x^2+x+1}{x^2}}=\dfrac{x^3-1}{x^2+x+1}\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
\(=\dfrac{\dfrac{x^3-1}{x^2}}{\dfrac{x^2+x+1}{x^2}}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
Giải phương trình: \(\dfrac{1}{\text{x^2-3x+2}}+\dfrac{1}{\text{x^2-5x+6}}+\dfrac{1}{\text{x^2-7x+12}}+\dfrac{1}{\text{x^2-9x+20}}=\dfrac{1}{15}\) có tập nghiệm là:
A. S={-1;5} B. S={11} C. S=\(\varnothing\) D. S={11;-5}
\(đkxđ:x\ne1;2;3;4;5\\ \Leftrightarrow\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}=\dfrac{1}{15}\\ \Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}=\dfrac{1}{15}\\ \Leftrightarrow\dfrac{1}{x-5}-\dfrac{1}{x-1}=\dfrac{1}{15}\\ \Leftrightarrow60=x^2-6x+5\\ \)
\(\Leftrightarrow60=x^2-6x+5\\ \Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\\ \Rightarrow D\)
Giá trị của biểu thức A = \(\dfrac{1}{\text{1 x 2}}\) + \(\dfrac{1}{\text{2 x 3}}\) + \(\dfrac{1}{\text{3 x 4}}\) là
A = 1/1 - 1/2 + 1/3 - 1/3 + 1/4
A = 1/1 - 1/4
A = 3/4
vậy A = 3/4
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}\)
\(=1-\dfrac{1}{4}=\dfrac{4-1}{4}=\dfrac{3}{4}\)
\(A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}\)
\(=1-\dfrac{1}{4}\)
\(=\dfrac{3}{4}\)
| #Sahara |
1 a..Rút gọn biểu thức A = \(\dfrac{\text{ x 2 − 4 x + 4}}{\text{x 3 − 2 x 2 − ( 4 x − 8 ) }}\)
b. Rút gọn biểu thức B = \(\left(\dfrac{x+2}{\text{x }\sqrt{\text{x }}+1}-\dfrac{1}{\sqrt{\text{x}}+1}\right).\dfrac{\text{4 }\sqrt{x}}{3}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
