Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)
=\(\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\left(\sqrt{a}-4\right).\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
Điều kiện bạn tự ghi nhé 
\(B=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}+\dfrac{\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\left(\sqrt{a}+3\right).\left(\sqrt{a}-3\right)-\left(\sqrt{a}-2\right).\left(\sqrt{a}+2\right)+\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{a-9-a+4+\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{1}{\sqrt{a}+1}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\)
\(VT=\dfrac{1+cos2x}{cos2x}\times\dfrac{1+cos4x}{sin4x}\) (*)
Ta có: theo công thức hạ bậc có: \(cos^2x=\dfrac{1+cos2x}{2}\Leftrightarrow1+cos2x=2cos^2x\) (1)
Ta có: \(cos2x=1-sin^2x\Rightarrow cos4x=1-2sin^22x\) (2)
Tương Tự có \(sin2x=2sinx\times cosx\Rightarrow sin4x=2sin2x\times cos2x\) (3)
Thay (1),(2),(3) vào (*) ta được: \(VT=\dfrac{2cos^2x}{cos2x}\times\dfrac{1+\left(1-2sin^22x\right)}{2sin2x\times cos2x}\)
\(VT=\dfrac{2cos^2x\times2\left(1-sin^22x\right)}{cos^22x\times2sin2x}\) mà \(1-sin^22x=cos^22x\)
\(\Rightarrow VT=\dfrac{2cos^2x\times cos^22x}{cos^22x\times2sinx\times cosx}=\dfrac{cosx}{sinx}=tanx\left(đpcm\right)\)
đoạn cuối nhầm nha \(VT=\dfrac{cosx}{sinx}=cotx\left(đpcm\right)\)
Câu a hạ bậc rồi áp dụng cosa + cosb
Câu b thì mối liên hệ giữa tan với cot là ra
Thay = x ; 
là y nhé bạn =='.
Theo đề bài ta có :
\(\left\{{}\begin{matrix}x+y=23\\x\cdot y=132\\y-x=1\end{matrix}\right.\left(ĐK:x,y>0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=23-y\\x\cdot y=132\\y-\left(23-y\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=23-y\\x\cdot y=132\\2y=24\Rightarrow y=12\end{matrix}\right.\)
Thay y = 12 vào hai đẳng thức trên ta được :
\(x+12=23\Rightarrow x=11\) hay \(x\cdot12=132\Rightarrow x=11\)
Vậy \(\left\{{}\begin{matrix}x=11\\y=12\end{matrix}\right.\) hay \(=11\); \(=12\).
Lời giải:
Áp dụng bất đẳng thức AM-GM:
\(a^2+2=(a^2+1)+1\geq 2\sqrt{a^2+1}\)
Do đó mà \(\frac{a^2+2}{\sqrt{a^2+1}}\geq \frac{2\sqrt{a^2+1}}{\sqrt{a^2+1}}=2\) (đpcm)
Dấu bằng xảy ra khi \(a^2+1=1\Leftrightarrow a=0\)
Mk ghi lộn đề rùi
bài 110 sgk trang 49 toán lop 6. Xl nhá
ĐKXĐ:\(x\ge0;y\ge1;z\ge2\)
\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{x+y+z}{2}\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1+2\sqrt{y-1}+1\right)+\left(z-2+2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-2\right)^2=0\)
Mà \(\left\{\begin{matrix}\left(\sqrt{x-1}-1\right)^2\ge0\\\left(\sqrt{y-1}-1\right)^2\ge0\\\left(\sqrt{z-2}-2\right)^2\ge0\end{matrix}\right.\)\(\forall x;y;z\)
\(\Rightarrow\left\{\begin{matrix}\left(\sqrt{x-1}-1\right)^2=0\\\left(\sqrt{y-1}-1\right)^2=0\\\left(\sqrt{z-2}-2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-1}-1=0\\\sqrt{z-2}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x-1=1\\y-1=1\\z-2=4\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x=2\\y=2\\z=6\end{matrix}\right.\)
=> x02 + y02 + z02 = 22 + 22 + 62 = 44
\(A=\left(\dfrac{1+\sqrt{1-a}}{1-a+\sqrt{1-a}}+\dfrac{1+\sqrt{1+a}}{1+a+\sqrt{1+a}}\right)\cdot\dfrac{a^2-1}{2}+1\)
\(=\left(\dfrac{1+\sqrt{1-a}}{\sqrt{1-a}\cdot\left(\sqrt{1-a}+1\right)}+\dfrac{1+\sqrt{1+a}}{\sqrt{1-a}\cdot\left(\sqrt{1+a}+1\right)}\right)\cdot\dfrac{a^2-1}{2}+1\)
\(=\left(\dfrac{1}{\sqrt{1-a}}+\dfrac{1}{\sqrt{1+a}}\right)\cdot\dfrac{a^2-1}{2}+1\)
\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{\left(1-a\right)\cdot\left(1+a\right)}}\cdot\dfrac{a^2-1}{2}+1\)
\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1-a^2}}\cdot\dfrac{a^2-1}{2}+1\)
cảm ơn bạn nhiu nha