Tim gia tri lon nhat :
a, A= 5-| 2x - 1 |
b, B= \(\dfrac{1}{\left|x-1\right|+3}\)
c, C= x +\(\dfrac{1}{2}-\left|x-\dfrac{2}{3}\right|\)
tim gia tri lon nhat :
D=\(\dfrac{2\left|x\right|+3}{3\left|x\right|-1}\)
\(D=\dfrac{2\left|x\right|+3}{3\left|x\right|-1}\)
\(\left\{{}\begin{matrix}\left|x\right|\ge0\Rightarrow2\left|x\right|\ge0\Rightarrow2\left|x\right|+3\ge3\\\left|x\right|\ge0\Rightarrow3\left|x\right|\ge0\Rightarrow3\left|x\right|-1\ge-1\end{matrix}\right.\)
\(MAX_D\Rightarrow D\in Z^+\)
Chắc chắn: \(2\left|x\right|+3\in Z^+\)
\(\Rightarrow3\left|x\right|-1\in Z^+\)
\(MAX_D\Rightarrow MIN_{3\left|x\right|-1}\)
\(\Rightarrow3\left|x\right|-1=1\)
\(\Rightarrow3\left|x\right|=2\Rightarrow\left|x\right|=\dfrac{2}{3}\)
\(\Rightarrow MAX_D=\dfrac{2\left|\dfrac{2}{3}\right|+3}{3\left|\dfrac{2}{3}\right|-1}=\dfrac{\dfrac{13}{3}}{1}=\dfrac{13}{3}\)
a)\(\dfrac{2}{x+2}-\dfrac{1}{x+3}+\dfrac{2x+5}{\left(x+2\right)\left(x+3\right)}\)
b)\(\dfrac{2}{x+1}-\dfrac{1}{x+5}+\dfrac{2x+6}{\left(x+5\right)\left(x+1\right)}\)
c)\(\dfrac{-6}{x^2-9}-\dfrac{1}{x+3}+\dfrac{3}{x-3}\)
d)\(\dfrac{x}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\)
Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
Cho bieu thuc: \(Q=\left(\dfrac{x^2-2x}{2x^2+8}+\dfrac{2x^2}{x^2.\left(x-2\right)}\right).\left(\dfrac{x^2-x-2}{x^2}\right)\)
a, Rut gon bieu thuc Q
b, Tim gia tri ca x de Q co gia tri bang \(\dfrac{1}{4}\)
Cho 2 bieu thuc :
A=\(\dfrac{x-3}{x+2}va\) B= \(\dfrac{3}{x+3}+\dfrac{2}{x-3}-\dfrac{3x-9}{x^2-9}\left(x-2,x\ne3x\ne-3\right)\)
a, Tinh gia tri bieu thuc A khi x=5
b, Chung minh : B=\(\dfrac{2}{x-3}\)
c, Biet C = A.B, Tim x de c = \(\dfrac{-1}{3}\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
Giải các phương trình sau:
1. \(a,\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{2x-6}\)
\(b,\dfrac{1}{x-2}+\dfrac{5}{x+1}=\dfrac{3}{2-x}\)
\(c,\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
2. \(a,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(b,2x^2-6x+1\)
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
tim gia tri lon nhat cua bieu thuc \(f\left(x\right)=\dfrac{1}{x^4-x^2+1}\)
Ta co :\(\dfrac{1}{f\left(x\right)}=\) \(x^4-x^2+1=x^4-2.\dfrac{1}{2}x^2+\dfrac{1}{4}+\dfrac{3}{4}\)
= \(\left(x^2-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
=> f(x) ≤ \(\dfrac{4}{3}\)
Vay max f(x) =\(\dfrac{4}{3}\)
Thực hiên phép tính:
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
b) \(\dfrac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
d) \(\left[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right]:\dfrac{x-y}{x}\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
1, tính
a,\(\dfrac{x+1}{x+2}:\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\)
b,\(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
c,\(\dfrac{x+3}{x+1}-\dfrac{2x-1}{x-1}-\dfrac{x-3}{x^2-1}\)
d,\(\dfrac{bc}{\left(a-b\right).\left(a-c\right)}+\dfrac{ac}{\left(b-a\right).\left(b-c\right)}+\dfrac{ab}{\left(c-a\right).\left(a-b\right)}\)
a: \(=\dfrac{x+1}{x+2}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+1}{x+3}=\dfrac{\left(x+1\right)^2}{\left(x+2\right)^2}\)
b: \(=\dfrac{x+1}{x+2}:\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+1\right)\left(x+2\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x-3-2x^2-2x+x+1-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+1}{\left(x-1\right)\left(x+1\right)}=-1\)