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\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
Lời giải:
a)
Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)
Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)
\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)
\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)
b)
ĐK: \(x\geq 0; x\neq 4\)
\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)
\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)
\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c) Thêm ĐK: \(x\geq 1\)
Từ biểu thức P vừa tìm được:
\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)
Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)
\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :
\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)
Vậy..........
Câu 2-Ta có x^2+y^2=5
(x+y)^2-2xy=5
Đặt x+y=S. xy=P
S^2-2P=5
P=(S^2-5)/2
Ta lại có P=x^3+y^3=(x+y)^3-3xy(x+y)=S^3-3SP=S^3-3S(S^2-5)/2
Rùi tự tính
Câu1
Ta có P<=a+a/4+b+a/12+b/3+4c/3 (theo bdt cô sy)
=> P<=4/3(a+b+c)=4/3
Vậy Max p =4/3 khi a=4b=16c
1)
Điều kiện: \(x\geq \frac{-1}{2}\)
Bình phương hai vế:
\(x^2+4=(2x+1)^2=4x^2+4x+1\)
\(\Leftrightarrow 3x^2+4x-3=0\)
\(\Leftrightarrow x=\frac{-2\pm \sqrt{13}}{3}\)
Do \(x\geq -\frac{1}{2}\Rightarrow x=\frac{-2+\sqrt{13}}{3}\) là nghiệm duy nhất của pt.
2)
a) \(x^2+x+12\sqrt{x+1}=36\) (ĐK: \(x\geq -1\) )
\(\Leftrightarrow (x^2+x-12)+12(\sqrt{x+1}-2)=0\)
\(\Leftrightarrow (x-3)(x+4)+\frac{12(x-3)}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow (x-3)\left[x+4+\frac{12}{\sqrt{x+1}+2}\right]=0\)
Do \(x\geq -1\Rightarrow x+4+\frac{12}{\sqrt{x+1}+2}\geq 3+\frac{12}{\sqrt{x+1}+2}>0\)
Do đó \(x-3=0\Leftrightarrow x=3\) (thỏa mãn)
Vậy pt có nghiệm x=3
b) Đặt \(\left\{\begin{matrix} \sqrt{x^2+7}=a\\ x+4=b\end{matrix}\right.\)
PT tương đương:
\(x^2+7+4(x+4)-16=(x+4)\sqrt{x^2+7}\)
\(\Leftrightarrow a^2+4b-16=ab\)
\(\Leftrightarrow (a-4)(a+4)-b(a-4)=0\)
\(\Leftrightarrow (a-4)(a+4-b)=0\)
+ Nếu \(a-4=0\Leftrightarrow \sqrt{x^2+7}=4\Leftrightarrow x^2=9\Leftrightarrow x=\pm 3\) (thỏa mãn)
+ Nếu \(a+4-b=0\Leftrightarrow a=b-4\)
\(\Leftrightarrow \sqrt{x^2+7}=x\)
\(\Rightarrow x\geq 0\). Bình phương hai vế thu được: \(x^2+7=x^2\Leftrightarrow 7=0\) (vô lý)
Vậy pt có nghiệm \(x=\pm 3\)
Câu 3:
Ta có \(M=\frac{x^2+2000x+196}{x}\)
\(\Leftrightarrow M=x+2000+\frac{196}{x}\)
Áp dụng BĐT AM-GM ta có: \(x+\frac{196}{x}\geq 2\sqrt{196}=28\)
\(\Rightarrow M=x+\frac{196}{x}+2000\geq 28+2000=2028\)
Vậy M (min) =2028. Dấu bằng xảy ra khi \(\left\{\begin{matrix} x=\frac{196}{x}\\ x>0\end{matrix}\right.\Rightarrow x=14\)