ms thi cấp 3 xong đúng ko??  có bài nào kt kì 1 thì cho mk nhé

cần CM:

\(\dfrac{1}{S_{ABC}}+\dfrac{1}{S_{IBC}}=\dfrac{1}{S_{MBC}}+\dfrac{1}{S_{NBC}}\)

\(\Leftrightarrow1+\dfrac{S_{ABC}}{S_{IBC}}=\dfrac{S_{ABC}}{S_{MBC}}+\dfrac{S_{ABC}}{S_{NBC}}\)

\(\Leftrightarrow1+\dfrac{S_{ABC}}{S_{IBC}}=\dfrac{AB}{MB}+\dfrac{AC}{NC}\)

mới nghĩ đc tới đây thôi để mai nghĩ nốt nhé

\(\dfrac{1}{\sqrt{3x^2-7x+20}}=\dfrac{1}{\sqrt{3\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{12}}}>0\forall x\)

We have \(3x^2-7x+20=\dfrac{1}{12}\left(36x^2-84x+240\right)\) \(=\dfrac{1}{12}\left[\left(6x\right)^2-2.6x.7+49+191\right]\) \(=\dfrac{1}{12}\left(6x-7\right)^2+\dfrac{191}{12}\)

Because \(\dfrac{1}{12}\left(6x-7\right)^2\ge0\) \(\Leftrightarrow\dfrac{1}{12}\left(6x-7\right)^2+\dfrac{191}{12}\ge\dfrac{191}{12}>0\) or we have \(3x^2-7x+20>0\) whatever the real number \(x\) is. Therefore, \(\dfrac{1}{\sqrt{3x^2-7x+20}}\) is always deterministic for all real numbers \(x\).

\(=\dfrac{\sqrt{5}}{5}-\dfrac{3}{10}.4\sqrt{5}-2\sqrt{5}+\sqrt{5}=\)

\(=\dfrac{\sqrt{5}}{5}-\dfrac{6\sqrt{5}}{5}-\sqrt{5}=-2\sqrt{5}\)

lạp lạp lạp

\(\dfrac{3}{2}\)\(\sqrt{x+2}\) - \(\sqrt{x+2}\) = \(\dfrac{4}{5}\)   đk x ≥ -2

\(\sqrt{x+2}\) ( \(\dfrac{3}{2}\) - 1) = \(\dfrac{4}{5}\)

\(\sqrt{x+2}\) . \(\dfrac{1}{2}\) = \(\dfrac{4}{5}\)

\(\sqrt{x+2}\)= \(\dfrac{4}{5}\) : \(\dfrac{1}{2}\)

\(\sqrt{x+2}\) = \(\dfrac{8}{5}\)

x + 2 = \(\dfrac{64}{25}\)

 x =  \(\dfrac{64}{25}\) - 2

x = \(\dfrac{14}{25}\)

= 3,23197531

= - 5,022646212216401

1/ \(C=a^2+\dfrac{18}{a}=\left(\dfrac{a^2}{24}+\dfrac{9}{a}+\dfrac{9}{a}\right)+\dfrac{23}{24}a^2\ge3\sqrt[3]{\dfrac{a^2}{24}.\dfrac{9}{a}.\dfrac{9}{a}}+\dfrac{23}{24}.6^2=3.\dfrac{3}{2}+\dfrac{69}{2}=39\)

Dấu "=" xảy ra \(\Leftrightarrow x=6\)

Vậy \(MinC=39\)

3/ \(A=\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}=\sum\limits_{cyc}\dfrac{1}{a^2+1}=\sum\limits_{cyc}\left(1-\dfrac{a^2}{a^2+1}\right)\ge\sum\limits_{cyc}\left(1-\dfrac{a^2}{2a}\right)=\sum\limits_{cyc}\left(1-\dfrac{a}{2}\right)=1-\dfrac{a}{2}+1-\dfrac{b}{2}+1-\dfrac{c}{2}=3-\dfrac{a+b+c}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Vậy \(MinA=\dfrac{3}{2}\)

a) We have \(3\le x\le5\)

Suppose \(\sqrt{5-x}=a\left(a\ge0\right);\sqrt{x-3}=b\left(b\ge0\right)\)

We imidiately have \(a+b=\sqrt{2}\)

On the other hand, we have \(a^2-b^2=\left(\sqrt{5-x}\right)^2-\left(\sqrt{x-3}\right)^2=\left(5-x\right)-\left(x-3\right)=8\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=8\) \(\Leftrightarrow\sqrt{2}\left(a-b\right)=8\) \(\Leftrightarrow a-b=4\sqrt{2}\)

And now we simply have \(\left\{{}\begin{matrix}a+b=\sqrt{2}\\a-b=4\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\sqrt{2}}{2}\left(take\right)\\b=\dfrac{3\sqrt{2}}{2}\left(take\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{5-x}=\dfrac{5\sqrt{2}}{2}\\\sqrt{x-3}=\dfrac{3\sqrt{2}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-x=\dfrac{25}{2}\\x-3=\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{2}\left(eliminate\right)\\x=\dfrac{15}{2}\left(take\right)\end{matrix}\right.\)

Therefore, this equation have the root \(x=\dfrac{15}{2}\)

b) We have \(x\ge2\)

\(\sqrt{x^2-4}=2\sqrt{x-2}\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-2\sqrt{x-2}=0\) \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\left(take\right)\)

So this equation has the root \(x=2\)

c) We have \(x\ge-\dfrac{4}{3}\). Suppose \(2x+3=A\left(A\ge0\right);3x+4=B\left(B\ge0\right)\). Notice that \(A+B=2x+3+3x+4=5x+7\). Thus, we can rewrite the equation as below:

\(\sqrt{A+B}=\sqrt{A}+\sqrt{B}\) (1)

Remember that \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\left(A,B\ge0\right)\). We can prove it easily by squaring each side of this inequality:

\(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\Leftrightarrow\left(\sqrt{A}+\sqrt{B}\right)^2\ge\left(\sqrt{A+B}\right)^2\)\(\Leftrightarrow A+B+2\sqrt{AB}\ge A+B\Leftrightarrow2\sqrt{AB}\ge0\Leftrightarrow\sqrt{AB}\ge0\). This is always true when \(A,B\ge0\). "=" happens when one of the expression A, B is equal to 0.

Therefore, (1) happens when \(\left[{}\begin{matrix}A=0\\B=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(eliminate\right)\\x=-\dfrac{4}{3}\left(take\right)\end{matrix}\right.\)

Thus, this equation has the root \(x=-\dfrac{4}{3}\)

d) We have \(x\ge-\dfrac{1}{3}\)

We can see that \(x=-\dfrac{1}{11}\). So, \(\sqrt{5x+7}=\dfrac{6\sqrt{22}}{11}\); o\(\sqrt{x+3}=\dfrac{4\sqrt{22}}{11}\) and \(\sqrt{3x+1}=\dfrac{2\sqrt{22}}{11}\). Therefore, we can rewrite the equation as below:

\(\left(\sqrt{5x+7}-\dfrac{6\sqrt{22}}{11}\right)-\left(\sqrt{x+3}-\dfrac{4\sqrt{22}}{11}\right)-\left(\sqrt{3x+1}-\dfrac{2\sqrt{22}}{11}\right)=0\)Now you multiply and devide each of the terms by the conjugate expression. The first term will has \(5x+\dfrac{5}{11}=5\left(x+\dfrac{1}{11}\right)\) as the numerator, the second term's numerator will be \(x+\dfrac{1}{11}\), and the final term has \(3x+\dfrac{3}{11}=3\left(x+\dfrac{1}{11}\right)\) as the numerator. And now you can see the commom factor \(x+\dfrac{1}{11}\)

\(\left(\sqrt{A}-B=\dfrac{\left(\sqrt{A}-B\right)\left(\sqrt{A}+B\right)}{\sqrt{A}+B}=\dfrac{A-B^2}{\sqrt{A}-B}\right)\)

x³ + (x + 1)³ + (x + 2)³ 

= x³ + (x + 1)³ + (x + 2)³ - 3x(x + 1)(x + 2) + 3x(x + 1)(x + 2)

= \(\dfrac{\left(x+x+1+x+2\right)\left[\left(x-x-1\right)^2+\left(x-x-2\right)^2+\left(x+1-x-2\right)^2\right]}{2}+3x\left(x+1\right)\left(x+2\right)\)

\(=9\left(x+1\right)+3x\left(x+1\right)\left(x+2\right)⋮9\) (vì x(x + 1)(x + 2) tích 3 số nguyên liên tiếp)