\(\sqrt{5^{2^{ }}-4^2}\) = \(\sqrt{25-16}\) = \(\sqrt{9}\) = 3 

\(\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3\)

biến đổi thành -[(căn x-2) -1]²  -[(căn x+1) -2]²+24 nhé bạn rồi suy ra max s=24 vs x=3 nhá 

Điều kiện \(x\ge-1\)

pt đã cho \(\Leftrightarrow x^2+x=3\left(\sqrt{x^3+1}-1\right)\)   (1)

Vì \(\sqrt{x^3+1}+1\ne0\) với mọi \(x\ge-1\) nên ta có thể viết lại pt (1) như sau:

\(\left(1\right)\Leftrightarrow x^2+x=3.\dfrac{\left(\sqrt{x^3+1}-1\right)\left(\sqrt{x^3+1}+1\right)}{\sqrt{x^3+1}+1}\)

\(\Leftrightarrow x^2+x=3.\dfrac{\left(\sqrt{x^3+1}\right)^2-1}{\sqrt{x^3+1}+1}\) 

\(\Leftrightarrow x^2+x=3.\dfrac{x^3}{\sqrt{x^3+1}+1}\)

\(\Leftrightarrow x\left(x+1-\dfrac{x^2}{\sqrt{x^3+1}+1}\right)=0\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x+1-\dfrac{x^2}{\sqrt{x^3+1}+1}=0\left(\cdot\right)\end{matrix}\right.\)

Xin lỗi bạn nhưng mình chỉ làm được đến đó thôi. Tìm được \(x=0\) rồi. Còn \(\left(\cdot\right)\) thì mình chưa giải được.

Chỗ kia mình nhầm xíu. \(\left(\cdot\right)\) phải là \(x+1=\dfrac{3x^2}{\sqrt{x^3+1}+1}\)

mong các CTV giải giúp ạ!

giải nhé:

ĐK: ko cần.

\(PT\Leftrightarrow3\left(x-1\right)+4\sqrt{x^2+x+1}=5\sqrt{2x^2-x+2}\)

đặt:\(a=x-1;b=\sqrt{x^2+x+1}\left(b>0\right)\)

\(\Rightarrow2x^2-x+2=a^2+b^2\)

\(PT\Leftrightarrow3a+4b=5\sqrt{a^2+b^2}\)

\(\Leftrightarrow9a^2+16b^2+24ab=25a^2+25b^2\)

\(\Leftrightarrow16a^2-24ab+9b^2=0\)

\(\Leftrightarrow\left(4a-3b\right)^2=0\)

\(\Leftrightarrow4a=3b\)

\(\Leftrightarrow4.\left(x-1\right)=3.\sqrt{x^2+x+1}\)

\(\Leftrightarrow4\left(x^2-2x+1\right)=3\left(x^2+x+1\right)\)

\(\Leftrightarrow4x^2-8x+4-3x^2-3x-3=0\)

\(\Leftrightarrow x^2-11x+1=0\)

thế thôi nha

Bạn nên nhớ \(\sin^2\alpha+\cos^2\alpha=1\)

Lại thêm \(\cos\alpha-\sin\alpha=\dfrac{1}{5}\Leftrightarrow\sin\alpha=\cos\alpha-\dfrac{1}{5}\) nên ta có \(\left(\cos\alpha-\dfrac{1}{5}\right)^2+\cos^2\alpha=1\) \(\Leftrightarrow\cos^2\alpha-\dfrac{2}{5}\cos\alpha+\dfrac{1}{25}+\cos^2\alpha-1=0\)

\(\Leftrightarrow2\cos^2\alpha-\dfrac{2}{5}\cos\alpha-\dfrac{24}{25}=0\)

\(\Leftrightarrow\cos^2\alpha-\dfrac{1}{5}\cos\alpha-\dfrac{12}{25}=0\)

\(\Leftrightarrow25\cos^2\alpha-5\cos\alpha-12=0\)

Đặt \(\cos\alpha=p\left(0< p< 1\right)\) thì ta có \(25p^2-5p-12=0\)

Ta có \(\Delta=\left(-5\right)^2-4.25\left(-12\right)=1225>0\), vậy:

\(x_1=\dfrac{-\left(-5\right)+\sqrt{1225}}{2.25}=\dfrac{4}{5}\left(nhận\right)\)

\(x_2=\dfrac{-\left(-5\right)-\sqrt{1225}}{2.25}=-\dfrac{3}{5}\left(loại\right)\)

Vậy ta có \(\cos\alpha=\dfrac{4}{5}\). Ta lại có \(\sin\alpha=\cos\alpha-\dfrac{1}{5}=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

Mà \(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\). Vậy \(\cot\alpha=\dfrac{4}{3}\)

Bài toán quá hay (người ra đề quá đẳng cấp)

A = \(\dfrac{2020}{2019^2+1}\) + \(\dfrac{2020}{2019^2+2}\)+......+\(\dfrac{2020}{2019^{2^{ }}+2019}\)

A = 2020 x ( \(\dfrac{1}{2019^{2^{ }}+1}\)+ \(\dfrac{1}{2019^2+2}\)+....+\(\dfrac{1}{2019^2+2019}\))

đặtB =( \(\dfrac{1}{2019^{2^{ }}+1}\)+ \(\dfrac{1}{2019^2+2}\)+....+\(\dfrac{1}{2019^2+2019}\))⇒ A =2020.B

mặt khác ta có   \(\dfrac{1}{2019^2+1}\) > \(\dfrac{1}{2019^2+2}\)>.....>\(\dfrac{1}{2019^2+2019}\)

⇔\(\dfrac{2019}{2019^2+1}\) > \(\dfrac{1}{2019^{2^{ }}+1}\)> \(\dfrac{1}{2019^{2^{ }}+2}\)+......+\(\dfrac{1}{2019^2+2019}\) >  \(\dfrac{2019}{2019^{2^{ }}+2019}\)

      ⇔       \(\dfrac{2019}{2019^{2^{ }}+2019}\) < B < \(\dfrac{2019}{2019^{2^{ }}+1}\)

⇔            \(\dfrac{2020.2019}{2019^{2^{ }}+2019}\) <2020. B < \(\dfrac{2020.2019}{2019^{2^{ }}+1}\)

⇔   \(\dfrac{2019.2020}{2019.\left(2019+1\right)}\) < 2012.B < \(\dfrac{\left(2019+1\right).2019}{2019^{2^{ }}+1}\)

⇔   \(\dfrac{2019.2020}{2019.2020}\)< 2020.B < \(\dfrac{2019^{2^{ }}+2019}{2019^{2^{ }}+1}\) = 1 + \(\dfrac{2018}{2019^{2^{ }}+1}\)< 2

⇔ 1 < 2020 .B < 2

⇔ 1 < A < 2

⇔ A không phải là số nguyên điều phải chứng minh 

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 \(x\).