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1)\(ĐKXĐ:x\ne0\)
Đặt \(\left(x+\dfrac{1}{x}\right)^2=a\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=a-2\)
\(\Rightarrow VT=2a+\left(a-2\right)^2-\left(a-2\right)a\)
\(=2a+a^2-4a+4-a^2+2a=4\)
\(\Rightarrow\left(x+2\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=-4\end{matrix}\right.\)
a) Đúng
b)Đúng
c)Sai vì nghiệm không thỏa mãn ĐKXĐ
d)Sai vì có 1 nghiệm không thỏa mãn ĐKXĐ
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
áp dụng bđt cauchy-shwarz dạng engel
\(\text{ Σ}_{cyc}\frac{a^2}{b+c}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)\(=\frac{a+b+c}{2}\)
Ta có hđt \(\text{ Σ}_{cyc}a^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Mà a+b+c khác 0 nên a = b = c
\(\Rightarrow N=1\)
5.\(C\text{ó}x^2-12=0\Rightarrow x^2=12\Rightarrow x=\sqrt{12}ho\text{ặc}x=-\sqrt{12}\)
Mà x>0\(\Rightarrow x=\sqrt{12}\)
6.Vì x-y=4\(\Rightarrow\left(x-y\right)^2=x^2-2xy+y^2=x^2-10+y^2=4^2=16\Rightarrow x^2+y^2=26\)
Có \(\left(x+y\right)^2=x^2+2xy+y^2=26+10=36=6^2=\left(-6\right)^2\)
Vì xy>0 và x>0 =>y>0=>x+y>0=>x+y=6
7. \(3x^2+7=\left(x+2\right)\left(3x+1\right)\)
\(3x^2+7=3x^2+7x+2\)
\(3x^2+7-3x^2-7x-2=0\)
-7x+5=0
-7x=-5
\(x=\frac{5}{7}\)
8.\(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\left(2x+1\right)^2-\left(2x+4\right)^2=9\)
(2x+1-2x-4)(2x+1+2x+4)=9
-3(4x+5)=9
4x+5=-3
4x=-8
x=-2
Còn câu 9 và 10 để mình nghiên cứu đã
1/
\(y\left(x+1\right)-x^2\left(x+1\right)=7\Leftrightarrow\left(x+1\right)\left(y-x^2\right)=7\)
TH1: \(\left\{{}\begin{matrix}x+1=1\\y-x^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=7\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+1=7\\y-x^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=37\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x+1=-1\\y-x^2=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x+1=-7\\y-x^2=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=63\end{matrix}\right.\)
2/
\(\left(1+\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right)...\left(1+\dfrac{1}{\left(x+1-1\right)\left(x+1+1\right)}\right)=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{\left(x+1\right)^2}{x\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{2.3.4...\left(x+1\right)}{3.4.5...\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow2012\left(x+1\right)=2011\left(x+2\right)\)
\(\Leftrightarrow x=2010\)
Bài 1:
\(\Leftrightarrow\dfrac{5x^2+15x-\left(x+2\right)\left(x+3\right)}{x^2-9}=2\)
\(\Leftrightarrow\dfrac{5x^2+15x-\left(x^2+5x+6\right)-2x^2+18}{\left|x\right|\ne3}=0\)
\(\Leftrightarrow2x^2-10x+12=0\)
\(\left(x+2\right)\left(x+6\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\) khác +-3 => kết luận hai nghiệm
Bài 4: tìm nghiệm lớn => xét x>-21
\(x^2-2x+1+x+21-x^2-5=0\)
\(\Leftrightarrow x=-5+22=17\Rightarrow x=17\) >-21 nhận