Cho Pt \(\dfrac{x+a}{a-x}+\dfrac{x-a}{a+x}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
a) GPT với a = -3
b) Tìm a biết x=\(\dfrac{1}{2}\)
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
cho phương trình ẩn x :
\(\dfrac{x-a}{x+a}-\dfrac{x+a}{x-a}+\dfrac{3a^2+a}{x^2-a^2}=0\)
a/ GPT với a = -3
b/GPT với a= 1
c/ Xác định a để pt có nghiêm x=0,5
a: Khi a=-3 thì phương trình sẽ là:
\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}+\dfrac{3\cdot9-3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9+24=0\)
=>12x=-24
hay x=-2
b: Khi a=1 thì phương trình trở thành:
\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x^2-2x+1-x^2-2x-1+4=0\)
=>-4x+4=0
hay x=1(loại)
cho phương trình ẩn x :
\(\dfrac{x-a}{x+a}-\dfrac{x+a}{x-a}+\dfrac{3a^2+a}{x^2-a^2}\)
a/ GPT với a = -3
b/GPT với a= 1
c/ Xác định a để pt có nghiêm x=0,5
help!!!!!!!\(\Leftrightarrow A=\dfrac{\left(x-a\right)^2-\left(x+a\right)^2+3a^2+a}{\left(x-a\right)\left(x+a\right)}\)
\(\Leftrightarrow A=\dfrac{-4ax+3a^2+a}{\left(x-a\right)\left(x+a\right)}\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|\ne a\\4ax=a\left(3a+1\right)\left(1\right)\end{matrix}\right.\)
a) với a=-3
\(\left(1\right)\Leftrightarrow4x=3.\left(-3\right)+1\Rightarrow x=-2\)(NHAN)
b)với a=-1
\(\left(1\right)\Leftrightarrow4x=3.\left(-1\right)+1\Rightarrow x=-\dfrac{2}{4}=-\dfrac{1}{2}\)(NHẬN)
c)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\x=\dfrac{3a+1}{4}=0,5\Rightarrow a=\dfrac{1}{3}\left(nhan\right)\end{matrix}\right.\)
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 \(x=\dfrac{a}{3x+2}\)
Cho biểu thức \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
a) Rút gọn \(A\)
b) Tính \(A\) biết \(\left|x-3\right|=2\)
c) Tìm \(x\) để \(A=\dfrac{1}{2}\)
d) Tìm \(x\) để \(A>1\)
e) Tìm \(x\) nguyên để \(A\) có giá trị nguyên
f) Với \(x>1\). Tìm giá trị nhỏ nhất của \(A\).
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
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: \(=\dfrac{3x^2-x+3+1-2x+x^2-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2x^2-5x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
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: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)
cho phương trình (ẩn x) \(\dfrac{x+a}{a-x}\) - \(\dfrac{x-a}{a+x}\) = \(\dfrac{a\left(3a+1\right)}{a^2-x^2}\) tìm các giá trị của a sao cho phương trình nhận x=\(\dfrac{1}{2}\) làm nghiệm giúp mình với ạ
Đk:\(a\ne\pm x\)
Pt \(\Leftrightarrow\dfrac{\left(a+x\right)^2-\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{2\left(a^2+x^2\right)}{a^2-x^2}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow2a^2+2x^2=3a^2+a\)
\(\Leftrightarrow a^2+a-2x^2=0\) (1)
Thay \(x=\dfrac{1}{2}\) vào (1) ta được:
\(a^2+a-2\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow a^2+a-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{-1+\sqrt{3}}{2}\\a=\dfrac{-1-\sqrt{3}}{2}\end{matrix}\right.\) (tm)
Vậy...
Giải phương trình với tham số a:
\(3x+\dfrac{x}{a}-\dfrac{3a}{a+1}=\dfrac{4ax}{\left(a+1\right)^2}+\dfrac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\dfrac{3a^2}{\left(a+1\right)^3}\).