Tìm x: a, \(\dfrac{\left|x\right|}{6-a-a^2}=\dfrac{1}{a+3}+\dfrac{1}{2-a}\)
b, \(\dfrac{\left|x\right|-3}{a^2-3a+2}=\dfrac{1}{a-1}-\dfrac{1}{a-2}\)
a) \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
b) \(\dfrac{a+2b}{3a-b}+\dfrac{2a-5b}{b-3a}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
d) \(\dfrac{4x}{x^2-4}+\dfrac{x}{x+2}+\dfrac{2}{x-2}\)
e) \(\dfrac{3x^2-x+3}{x^3-1}+\dfrac{1-x}{x^2+x+1}+\dfrac{2}{1-x}\)
f) \(\dfrac{1}{x^2+3x+2}+\dfrac{1-x}{x^2+x+1}+\dfrac{2}{1-x}\)
g) \(\dfrac{a^3}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right)\left(c-b\right)}\)
h) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
b: \(=\dfrac{a+2b}{3a-b}-\dfrac{2a-5b}{3a-b}\)
\(=\dfrac{a+2b-2a+5b}{3a-b}=\dfrac{-a+7b}{3a-b}\)
c: \(=\dfrac{2+x-3}{\left(x+3\right)\left(x-3\right)}=\dfrac{x+1}{\left(x+3\right)\left(x-3\right)}\)
d: \(=\dfrac{4x+x^2-2x+2x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)
e: \(=\dfrac{3x^2-x+3+1-2x+x^2-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2x^2-5x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
Giải các pt với tham số là a,b,c
a , \(\dfrac{x-a}{3}=\dfrac{x+3}{a}-2\) e, \(3x+\dfrac{x}{a}-\dfrac{3a}{a+1}=\dfrac{4ax}{\left(a+1\right)^2}+\dfrac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\dfrac{3a^2}{\left(a+1\right)^3}\)
b, \(\dfrac{x-a}{a+1}+\dfrac{x-1}{a-1}=\dfrac{2a}{1-a^2}\)
c, \(\dfrac{x+a-1}{a+2}+\dfrac{x-a}{a-2}+\dfrac{x-a}{4-a^2}\)
d, \(\dfrac{x-a}{b+c}+\dfrac{x-b}{c+a}+\dfrac{x-c}{a+b}=3\)
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
g, x - a / b + c + x - b/ c+a + x - c/ a+b = 3x / a+b+c
Rút gọn:
\(A=\left[\left(\dfrac{3}{1+x}-\dfrac{x}{x^2+x+1}\right):\dfrac{2x^2+3x}{x+1}+\dfrac{3}{x+1}\right]\cdot\dfrac{x^2+x}{1+3x}\)
\(B=\left[\dfrac{a}{2a-6}-\dfrac{a^2}{a^2-9}+\dfrac{a}{2a-9}\cdot\left(\dfrac{3}{a}+\dfrac{1}{3-a}\right)\right]:\dfrac{a^2-5a-6}{18-2a^2}\)
Câu 1: Rút gọn
a) \(\left(1+\dfrac{a}{x}+\dfrac{a^2}{x^2}\right)\left(1-\dfrac{a}{x}\right)\left(\dfrac{x^2}{a^3-x^3}\right)\)
b) \(\dfrac{x+3a}{2-x}+\dfrac{x-3a}{2+x}-\dfrac{2a}{4-x^2}+a\)
Biết \(x=\dfrac{a}{3x+2}\)
Giải phương trình với tham số a:
\(3x+\dfrac{x}{a}-\dfrac{3a}{a+1}=\dfrac{4ax}{\left(a+1\right)^2}+\dfrac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\dfrac{3a^2}{\left(a+1\right)^3}\).
Bài 1: a;b;c > 0
Chứng minh : \(\dfrac{a}{3a+b+c}+\dfrac{b}{3b+a+c}+\dfrac{c}{3c+a+b}\le\dfrac{3}{5}\)
Bài 2: x;y;z \(\ne\) 1 và xyz = 1
Chứng minh : \(\dfrac{x^2}{\left(x-1\right)^2}+\dfrac{y^2}{\left(y-1\right)^2}+\dfrac{z^2}{\left(z-1\right)^2}\ge1\)
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)
Câu 1:
\(C=\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}\cdot\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{4-x^2}\right)\)
a) Rút gọn C
b) x bằng mấy để C = 1?
Câu 2:
\(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a) Rút gọn B
b) x bằng mấy để \(\left|B\right|=B\)
Câu 3: Rút gọn:
\(A=\left[\dfrac{\left(1-a\right)^2}{3a+\left(a-1\right)^2}+\dfrac{2a^2-4a-1}{a^3-1}-\dfrac{1}{1-a}\right]:\dfrac{2a}{a^3+a}\)
rút gọn biểu thức
1) \(\dfrac{a+b}{3a-b}+\dfrac{b}{a+b}-\dfrac{a^2-b^2}{3a-b}\)
2) \(\left(\dfrac{7}{a+b}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right)\div\dfrac{a+1}{2}\)
3) \(\left(x^2+\dfrac{4x^2}{x^2-4}\right)\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x^3-4x}\times\dfrac{x^2-4}{x-2}\right)\)
2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)
\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)
\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)
3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)
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.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)