Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)
c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(x-1\right)^2}{4x}\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
b.
\(\left|x-5\right|=4\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{9}+1}{2\sqrt{9}}=\dfrac{2}{3}\)
2)
\(A=\dfrac{5\sqrt{a}-3}{\sqrt{a}-2}+\dfrac{3\sqrt{a}+1}{\sqrt{a}+2}-\dfrac{a^2+2\sqrt{a}+8}{a-4}\)
\(=\dfrac{\left(5\sqrt{a}-3\right)\left(\sqrt{a}+2\right)+\left(3\sqrt{a}+1\right)\left(\sqrt{a}-2\right)-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}+\sqrt{a}-2-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{-a^2+8a-16}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{-\left(a-4\right)^2}{a-4}=4-a\)
1: Ta có: \(\left\{{}\begin{matrix}3x-y=2m-1\\x+y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=5m+1\\x+y=3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=3m+2-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=\dfrac{12m+8-5m-1}{4}=\dfrac{7m+7}{4}\end{matrix}\right.\)
Ta có: \(x^2+2y^2=9\)
\(\Leftrightarrow\left(\dfrac{5m+1}{4}\right)^2+2\cdot\left(\dfrac{7m+7}{4}\right)^2=9\)
\(\Leftrightarrow\dfrac{25m^2+10m+1}{16}+\dfrac{2\cdot\left(49m^2+98m+49\right)}{16}=9\)
\(\Leftrightarrow25m^2+10m+1+98m^2+196m+98-144=0\)
\(\Leftrightarrow123m^2+206m-45=0\)
Đến đây bạn tự làm nhé, chỉ cần giải phương trình bậc hai bằng delta thôi
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}
a/ Ta có
\(K^4+\frac{1}{4}=K^4+K^2+\frac{1}{4}-K^2=\left(K^2+\frac{1}{2}\right)^2-K^2=\left(K^2+K+\frac{1}{2}\right)\left(K^2-K+\frac{1}{2}\right)\)
Ta lại có
\(K^2+K+\frac{1}{2}=\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\)
\(\Rightarrow K^4+\frac{1}{4}=\left(K^2-K+\frac{1}{2}\right)\left(\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\right)\)
Áp dụng vào bài toán ta được
\(=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)\(1S=\frac{\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)\left(5^2-5+0,5\right)...\left(100^2-100+0,5\right)\left(101^2-101+0,5\right)}{\left(1^2-1+0,5\right)\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)...\left(99^2-99+0,5\right)\left(100^2-100+0,5\right)}\)
b/
\(\frac{3\left(x+y\right)}{3\sqrt{x\left(4x+5y\right)}+3\sqrt{y\left(4y+5x\right)}}\)
\(\ge\frac{3\left(x+y\right)}{\frac{9x+4x+5y}{2}+\frac{9y+4y+5x}{2}}\)
\(=\frac{1}{3}\)
Dấu = xảy ra khi x = y
\(a, A=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=\left(2-3-4\right)\sqrt{x-1}=-5\sqrt{x-1}\)
\(b, B=\frac{2}{x+y}.\left(x+y\right)\sqrt{\frac{3}{4}}=2\sqrt{\frac{3}{4}}=2.\frac{1}{2}.\sqrt{3}=\sqrt{3}\)
\(A=3x\left(x-4y\right)-\dfrac{12}{5}y\left(y-5x\right)\)
\(A=3x^2-12xy-\dfrac{12}{5}y^2+12xy\)
\(A=3x^2-\dfrac{12}{5}y^2\)
Thay \(x=4;y=-5\) vào A ta được:
\(3\cdot4^2-\dfrac{12}{5}\cdot\left(-5\right)^2=48-60=-12\)
Vậy ....