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T
1
AH
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AH
0
26 tháng 5 2021
Ta có
\(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right).\left(a+c\right)\\ Cmtt:b^2+1=\left(b+a\right).\left(b+c\right)\\ c^2+1=\left(c+a\right).\left(c+b\right)\)
Nên
\(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\\ =\dfrac{\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{\left(c-a\right)}{\left(b+c\right)\left(b+a\right)}+\dfrac{\left(a-b\right)}{\left(c+a\right)\left(c+b\right)}\\ =\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =0\)
26 tháng 5 2021
\(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\)
\(=\dfrac{b-c}{a^2+ab+bc+ac}+\dfrac{c-a}{b^2+ab+bc+ca}+\dfrac{a-b}{c^2+ab+bc+ca}\)
\(=\dfrac{b-c}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{c-a}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{a-b}{c\left(c+a\right)+b\left(a+c\right)}\)
\(=\dfrac{b-c}{\left(a+c\right)\left(a+b\right)}+\dfrac{c-a}{\left(b+c\right)\left(a+b\right)}+\dfrac{a-b}{\left(b+c\right)\left(a+c\right)}\)
\(=\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(a+c\right)+\left(a-b\right)\left(a+b\right)}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)
\(=\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
2 tháng 11 2023
\(A=\dfrac{3}{2}-tana\cdot cos^2a\)
\(=\dfrac{3}{2}-\dfrac{sina}{cosa}\cdot cos^2a\)
\(=\dfrac{3}{2}-sina\cdot cosa\)
\(=\dfrac{3}{2}-\dfrac{1}{2}sin2a\)
\(0^0< a< 90^0\)
=>\(0< =2a< =180^0\)
=>\(sin2a\in\left[-1;1\right]\)
\(-1< =sin2a< =1\)
=>\(\dfrac{1}{2}>=-\dfrac{1}{2}sin2a>=-\dfrac{1}{2}\)
=>\(\dfrac{7}{2}>=-\dfrac{1}{2}sin2a+3>=\dfrac{5}{2}\)
=>\(\dfrac{5}{2}< =y< =\dfrac{7}{2}\)
\(y_{min}=\dfrac{5}{2}\) khi sin2a=1
=>\(2a=\dfrac{\Omega}{2}+k2\Omega\)
=>\(a=\dfrac{\Omega}{4}+k\Omega\)
mà 0<a<90
nên a=45
Y
6 tháng 6 2023
\(A=\left(\dfrac{\sqrt{x}-2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}+2}{x-1}\right):\dfrac{2\sqrt{x}}{x-1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}^2+2\sqrt{x}+1^2}-\dfrac{\sqrt{x}+2}{\sqrt{x}^2-1^2}\right).\dfrac{x-1}{2\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{x-1}{2\sqrt{x}}\)
Tới đây là có được mẫu chung ở dấu = thứ 2 rồi.
LD
6 tháng 6 2023
\(A=\left(\dfrac{\sqrt{x}-2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}+2}{x-1}\right):\dfrac{2\sqrt{x}}{x-1}\) ( với x>0;\(x\ne1\) )
\(=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\dfrac{x-1}{2\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}.\dfrac{x-1}{2\sqrt{x}}\)
\(=.....\) ( theo như trên )
21 tháng 8 2021
1: \(\dfrac{x^2-5}{x-\sqrt{5}}=x+\sqrt{5}\)
2: \(\dfrac{1-b\sqrt{b}}{1-\sqrt{b}}=1+\sqrt{b}+b\)
3: \(\dfrac{1-\sqrt{8}}{1+\sqrt{2}}=-5+3\sqrt{2}\)
Trả lời:
a, \(\Delta=b^2-4ac=\left(-5\right)^2-4.2.1=17>0\)
=> pt có 2 nghiệm phân biệt
Áp dụng hệ thức Vi-ét, ta có: \(\hept{\begin{cases}x_1+x_2=\frac{-b}{a}=\frac{-\left(-5\right)}{2}=\frac{5}{2}\\x_1x_2=\frac{c}{a}=\frac{1}{2}\end{cases}}\) (*)
b, \(A=3x_1^2+3x_2^2-5x_1x_2+7=3\left(x_1^2+x_2^2\right)-5x_1x_2+7\)
\(=3\left(x_1^2+2x_1x_2+x_2^2-2x_1x_2\right)-5x_1x_2+7\)
\(=3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2+7\) (1)
Thay (*) vào (1), ta được:
\(A=3\left[\left(\frac{5}{2}\right)^2-2\cdot\frac{1}{2}\right]-5\cdot\frac{1}{2}+7=\frac{81}{4}\)
c, \(B=4x_1+4x_2-8x_1^2-8x_2^2-5=4\left(x_1+x_2\right)-8\left(x_1^2+x_2^2\right)-5\)
\(=4\left(x_1+x_2\right)-8\left(x_1^2+2x_1x_2+x_2^2-2x_1x_2\right)-5\)
\(=4\left(x_1+x_2\right)-8\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5\) (2)
Thay (*) vào (2), ta được:
\(B=4\cdot\frac{5}{2}-8\left[\left(\frac{5}{2}\right)^2-2\cdot\frac{1}{2}\right]-5=-37\)
d, \(C=2x_1^3+2x_2^3+5=2\left(x_1^3+x_2^3\right)+5\)
\(=2\left(x_1^3+3x_1^2x_2+3x_1x_2^2+x_2^3-3x_1^2x_2-3x_1x_2^2\right)+5\)
\(=2\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]+5\) (3)
Thay (*) vào (3), ta được:
\(C=2\left[\left(\frac{5}{2}\right)^3-3\cdot\frac{1}{2}\cdot\frac{5}{2}\right]+5=\frac{115}{4}\)