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a) Phương trình có nghiệm bằng 1 khi \(1+a-4-4=0\)
\(\Rightarrow a=7\)
b) Khi a = 7 thì phương trình trở thành \(x^3+7x^2-4x-4=0\)
\(\Leftrightarrow-x^3-7x^2+4x+4=0\)
\(\Leftrightarrow\left(-x^3-8x^2-4x\right)+\left(x^2+8x+4\right)=0\)
\(\Leftrightarrow-x\left(x^2+8x+4\right)+\left(x^2+8x+4\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x^2+8x+4\right)=0\)
+) 1 - x = 0 thì x = 1
+) \(x^2+8x+4=0\)
\(\Leftrightarrow x^2+8x+16-12=0\Leftrightarrow\left(x+4\right)^2=12\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=\sqrt{12}\\x+4=-\sqrt{12}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{12}-4\\x=-\sqrt{12}-4\end{cases}}\)
Vậy phương trình có 3 nghiệm \(\left\{1;\pm\sqrt{12}-4\right\}\)
a) ĐKXĐ : \(x\ne\pm a\).
Với \(a=-3\) khi đó ta có pt :
\(A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left(-9+1\right)}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(-3-x\right)}{\left(-3-x\right)\left(-3+x\right)}+\frac{24}{\left(-3-x\right)\left(-3+x\right)}=0\)
\(\Rightarrow x^2-9-\left(-3x-x^2-9-3x\right)+24=0\)
\(\Leftrightarrow2x^2+6x+24=0\)
\(\Leftrightarrow x^2+3x+12=0\) ( vô nghiệm )
Phần b) tương tự.
\(A=\frac{x+a}{a-x}-\frac{x-a}{a+x}=\frac{a\left(3x+1\right)}{a^2-x^2}\)
\(=\frac{x+a}{a-x}+\frac{x-a}{a+x}=\frac{a\left(3+1\right)}{\left(a-x\right)\left(a+x\right)}\)
\(=\frac{\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+1\right)}=\frac{a\left(3a+1\right)}{\left(a+x\right)\left(a-x\right)}\)
\(\Leftrightarrow\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)=a\left(3a+1\right)\)
\(\Leftrightarrow x^2+2ax+a^2-ax-x^2-a^2+ax=3a^2+a\)
\(\Leftrightarrow2ax=3a^2+a\)
\(\Leftrightarrow x=\frac{3a^2+a}{2a}\left(a\ne0\right)\)
a) Khi x=-3 => \(x=\frac{3\cdot\left(-3\right)^2-3}{2\left(-3\right)}=-13\)
b) a=1
\(\Leftrightarrow x=\frac{3\cdot1^2+1}{2\cdot1}=2\)
a) \(ĐKXĐ:x\ne\pm3\)
Với a = -3
\(\Leftrightarrow A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}=\frac{24}{9-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}+\frac{24}{x^2-9}=0\)
\(\Leftrightarrow\frac{-\left(x-3\right)^2-\left(x+3\right)^2+24}{x^2-9}=0\)
\(\Leftrightarrow-x^2+6x-9-x^2-6x-9+24=0\)
\(\Leftrightarrow-2x^2+6=0\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\pm\sqrt{3}\)(tm)
Vậy với \(a=-3\Leftrightarrow x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
b) \(ĐKXĐ:x\ne\pm1\)
Với a = 1
\(\Leftrightarrow A=\frac{x+1}{1-x}-\frac{x-1}{1+x}=\frac{3+1}{1-x^2}\)
\(\Leftrightarrow\frac{x+1}{1-x}-\frac{x-1}{1+x}+\frac{4}{x^2-1}=0\)
\(\Leftrightarrow\frac{-\left(x+1\right)^2-\left(x-1\right)^2+4}{x^2-1}=0\)
\(\Leftrightarrow-x^2-2x-1-x^2+2x-1+4=0\)
\(\Leftrightarrow-2x^2+2=0\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)(ktm)
Vậy với \(a=1\Leftrightarrow x\in\varnothing\)
c) \(ĐKXĐ:a\ne\pm\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào phương trình, ta đươc :
\(A=\frac{\frac{1}{2}+a}{a-\frac{1}{2}}-\frac{\frac{1}{2}-a}{a+\frac{1}{2}}=\frac{a\left(3a+1\right)}{a^2-\frac{1}{4}}\)
\(\Leftrightarrow\frac{a+\frac{1}{2}}{a-\frac{1}{2}}+\frac{a-\frac{1}{2}}{a+\frac{1}{2}}-\frac{3a^2+a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow\frac{\left(a+\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2-3a^2-a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow a^2+a+\frac{1}{4}+a^2-a+\frac{1}{4}-3a^2-a=0\)
\(\Leftrightarrow-a^2-a+\frac{1}{2}=0\)
\(\Leftrightarrow a^2+a-\frac{1}{2}=0\)
\(\Leftrightarrow\left(a+\frac{1}{2}\right)^2-\frac{3}{4}=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}\\a=-\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{-\sqrt{3}-1}{2}\end{cases}}\)(TM)
Vậy với \(x=\frac{1}{2}\Leftrightarrow a\in\left\{\frac{\sqrt{3}-1}{2};\frac{-\sqrt{3}-1}{2}\right\}\)
\(\text{a) Thay a = 4 vào pt ta có:}\)
\(\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+4\right)+\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=2\)
\(\Leftrightarrow\frac{x^2-16+x^2-4}{x^2-4x+2x-8}=2\)
\(\Leftrightarrow\frac{2x^2-20}{x^2-2x-8}=2\)
\(\Leftrightarrow2x^2-20=2.\left(x^2-2x-8\right)\)
\(\Leftrightarrow2x^2-20=2x^2-4x-16\)
\(\Leftrightarrow2x^2-2x^2+4x=-16+20\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
\(\text{b) Thay x = -1 vào pt ta có:}\)
\(\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)
\(\Leftrightarrow\frac{a-1}{1}+\frac{-3}{-\left(a+1\right)}=2\)
\(\Leftrightarrow\left(a-1\right)+\frac{3}{a+1}=2\)
\(\Leftrightarrow\frac{\left(a-1\right)\left(a+1\right)+3}{a+1}=2\)
\(\Leftrightarrow\frac{a^2-1+3}{a+1}=2\)
\(\Leftrightarrow a^2+2=2.\left(a+1\right)\)
\(\Leftrightarrow a^2+2=2a+2\)
\(\Leftrightarrow a^2-2a=2-2\)
\(\Leftrightarrow a\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=0\\a-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
Vậy để pt có nghiệm là x = 1 thì a = {0 ; 2}
\(a.Thay:a=4\Leftrightarrow\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}=\frac{2\left(x+2\right)\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}\)
\(\Rightarrow\left(x+4\right)\left(x-4\right)+\left(x-2\right)\left(x+2\right)=2\left(x+2\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-4x+4x-16+x^2+2x-2x-4=\left(2x+4\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-20=2x^2-8x+4x-16\)
\(\Leftrightarrow2x^2-20-2x^2+8x-4x+16=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow x=1\)
a) Giải phương trình theo b khi a=3
Lời giải :
\(1-\dfrac{2b}{x-b}=\dfrac{a^2-b^2}{b^2+x^2-2bx}\)
\(\Leftrightarrow1-\dfrac{2b}{x-b}=\dfrac{a^2-b^2}{\left(b-x\right)^2}\)
\(\Leftrightarrow\) \(\dfrac{\left(x-b\right)^2}{\left(x-b\right)^2}-\dfrac{2b\left(x-b\right)}{\left(x-b\right)^2}=\dfrac{a^2-b^2}{\left(x-b\right)^2}\)
\(\Rightarrow\left(x-b\right)^2-2bx-2b^2=a^2-b^2\)
\(\Leftrightarrow x^2-2xb+b^2-2bx+2b^2=a^2-b^2\)
\(\Leftrightarrow x^2-2xb+b^2-2bx+2b^2-a^2+b^2=0\)
\(\Leftrightarrow x^2-4xb+4b^2-a^2=0\)
\(\Leftrightarrow\left(x-2b\right)^2-a^2=0\)
Tại a=3
=> \(a^2=9\)
\(\Leftrightarrow\left(x-2b\right)^2-9=0\)
\(\Leftrightarrow\left(x-2b-3\right)\left(x-2b+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2b-3=0\\x-2b+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2b=3\\x-2b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}b=\dfrac{x-3}{2}\\b=\dfrac{x+3}{2}\end{matrix}\right.\)
Akai Haruma
Ribi Nkok Ngok
Võ Đông Anh Tuấn
Gia Hân Ngô