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15
\(\dfrac{7}{x-2}\)+\(\dfrac{8}{x-5}\)=3 (x khác 2 khác 5)
\(\Leftrightarrow\)7*(x-5)+8(x-2)=3(x-2)(x-5)
\(\Leftrightarrow\)15x-51=3x^2-21x+30\(\Leftrightarrow\)3x^2-36x+81=0
\(\Leftrightarrow\)\(\begin{matrix}&\end{matrix}\)\(\left[{}\begin{matrix}9\\3\end{matrix}\right.\) tmđk
16\(\dfrac{x^2-3x+6}{x^2-9}\)=\(\dfrac{1}{x-3}\)(x khác +_3)
\(\Leftrightarrow\)x^2-3x+6=x+3
\(\Leftrightarrow\)x^2-4x+3=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}3loại\\1\end{matrix}\right.\)
vậy x=1 là nghiệm của pt
17 \(\dfrac{3}{x^2-4}\) = \(\dfrac{1}{x-2}+\dfrac{1}{x+2}\)
<=> x + 2 + x - 2 = 3
<=> 2x = 3
<=> x = \(\dfrac{3}{2}\)
a)\(\dfrac{2}{x^2-1}+\dfrac{1}{x+1}=2\) Điều kiện:x#1,-1
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x-1\right)}+\dfrac{1}{x+1}=2\\\)
\(\Leftrightarrow\dfrac{2+x-1}{\left(x+1\right)\left(x-1\right)}=2\)
\(\Leftrightarrow\dfrac{1}{x-1}=2\)
\(\Leftrightarrow1=2\left(x-1\right)\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
b)\(1-\dfrac{12}{x^2-4}=\dfrac{3}{x+2}\) Điều kiện:x#2,-2
\(\Leftrightarrow\dfrac{x^2-4-12}{x^2-4}=\dfrac{3}{x+2}\)
\(\Leftrightarrow x^2-16=3\left(x-2\right)\)
\(\Leftrightarrow x^2-16-3x+6=0\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{5\right\}\)
a) điều kiện xác định : \(x\ne\pm1\)
ta có : \(\dfrac{1}{x-1}-\dfrac{2}{x+1}=\dfrac{4}{x^2-1}\Leftrightarrow\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{3-x}{x^2-1}=\dfrac{4}{x^2-1}\Leftrightarrow3-x=4\Leftrightarrow x=-1\) vậy \(x=-1\)
câu này biến đổi xong nó ra luôn pt bật 1 nên o tính \(\Delta\) đc .
b) điều kiện xác định : \(-\sqrt{5}\le x\le\sqrt{5}\)
ta có : \(\sqrt{5-x^2}=x^2+1\Leftrightarrow5-x^2=x^4+2x^2+1\)
\(\Leftrightarrow x^4+3x^2-4=0\)
đặc \(x^2=t\left(t\ge0\right)\) \(\Rightarrow pt\Leftrightarrow t^2+3t-4=0\)
ta có : \(\Delta=3^2-4\left(-4\right)=9+16=25>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(t_1=\dfrac{-3+\sqrt{25}}{2}=1\) ; \(t_2=\dfrac{-3-\sqrt{25}}{2}=-4\left(loại\right)\)
với \(t=1\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(tmđk\right)\)
vậy \(x=\pm1\)
c) ta có : \(x^3-1=x^2-1\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2=0\end{matrix}\right.\) mấy cái này cũng o tính đen ta đc .
\(\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}\)
\(\Leftrightarrow\dfrac{\left(x-\dfrac{1}{x}\right)-\left(1-\dfrac{1}{x}\right)}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{x-1}{x}=0\)
\(\Leftrightarrow\dfrac{x-1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{x-1}{x}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{1}{x}\right)=0\)
Pt \(\dfrac{1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{1}{x}=0\) vô n0
=> x - 1 = 0
<=> x = 1 (nhận)
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
Lời giải:
ĐK: \(x\geq \frac{-1}{4}\)
Biến đổi:
\(x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=\left(x+\frac{1}{4}\right)+2.\frac{1}{2}\sqrt{x+\frac{1}{4}}+(\frac{1}{2})^2\)
\(=(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2\)
\(\Rightarrow \sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=\sqrt{x+\frac{1}{4}}+\frac{1}{2}\)
Do đó pt ban đầu tương đương với:
\(x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow (x+\frac{1}{4})+\sqrt{x+\frac{1}{4}}+\frac{1}{4}=2\)
\(\Leftrightarrow (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2=2\)
\(\Rightarrow \sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\) (TH bằng $-\sqrt{2}$ ta có thể loại luôn vì biểu thức không âm)
\(\Rightarrow x+\frac{1}{4}=(\sqrt{2}-\frac{1}{2})^2=\frac{9-4\sqrt{2}}{4}\)
\(\Rightarrow x=2-\sqrt{2}\) (t/m)
Vậy..............
\(\dfrac{1}{x}+\dfrac{1}{x+50}=\dfrac{1}{60}\left(x\ne0;x\ne-5\right)\)
\(pt\Leftrightarrow\dfrac{x+50}{x\left(x+50\right)}+\dfrac{x}{x\left(x+50\right)}=\dfrac{1}{60}\)
\(\Leftrightarrow\dfrac{2x+50}{x\left(x+50\right)}=\dfrac{1}{60}\Leftrightarrow x\left(x+50\right)=60\left(2x+50\right)\)
\(\Leftrightarrow x^2+50x=120x+3000\)
\(\Leftrightarrow x^2-70x-3000=0\)
\(\Leftrightarrow x^2-100x+30x-3000=0\)
\(\Leftrightarrow x\left(x-100\right)+30\left(x-100\right)=0\)
\(\Leftrightarrow\left(x+30\right)\left(x-100\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+30=0\\x-100=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-30\\x=100\end{matrix}\right.\)