Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH - Hỏi đáp

\(x+\sqrt{x-1}=13\)<=> \(\sqrt{x-1}=13-x\)

<=>\(\begin{cases}13-x\ge0\\x^2-27x+170=0\end{cases}\)

<=> \(\begin{cases}x\le13\\x=10,x=17\end{cases}\)

=> x=10

vậy nghiemj x=10

cho mình hỏi đề bài là:

2\(\sqrt{x}+1...\)hay là 2\(\sqrt{x+1}\)....

đk x\(\ge-1\)

pt \(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=25\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=25\)

\(\Leftrightarrow2\sqrt{2x^2+5x+3}=21-3x\)

\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(21-3x\right)^2\)đk \(x\le7\)

\(\Leftrightarrow8x^2+20x+12=9x^2-126x+441\)

\(\Leftrightarrow x^2-146x+429=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=143\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)

Áp dụng BĐT AM - GM, ta có:

\(a^2+2b^2+3\)

\(=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)

\(\ge2ab+2b+2\)

Tương tự, ta có: \(b^2+2c^2+3\ge2bc+2c+2\) và \(c^2+2a^2+3\ge2ac+2a+2\)

\(VT=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)

\(\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a^2bc+abc}\right)\) (Thay abc = 1)

\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1+ab+b}{ab+b+1}\)

\(=\dfrac{1}{2}=VP\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c = 1

a. 3xy + x - y = 1

<=> 9xy + 3x - 3y = 3

<=> 3x(3y+1) - (3y+1) = 2

<=> (3x-1)(3y+1)=2

Xét các trường hợp ta có x = 1, y = 0

Vậy nghiệm của pt là (1;0) ; (0;-1)

Đặt \(a=\sqrt{1-x^2},b=x\Rightarrow a^2+b^2=1-x^2+x^2=1\).
Ta có \(\dfrac{1}{a}+\dfrac{1}{b}=2\Leftrightarrow\dfrac{a+b}{ab}=2\sqrt{2}\) \(\Leftrightarrow\dfrac{a^2+b^2+2ab}{a^2b^2}=8\) \(\Leftrightarrow\dfrac{1+2ab}{a^2b^2}=8\)
Suy ra \(8a^2b^2-2ab-1=0\) \(\Leftrightarrow\left[{}\begin{matrix}ab=-\dfrac{1}{4}\\ab=\dfrac{1}{2}\end{matrix}\right.\).
Với \(ab=-\dfrac{1}{4}\)ta có: \(\left\{{}\begin{matrix}ab=-\dfrac{1}{4}\\a^2+b^2=1\end{matrix}\right.\)
.....
Với \(ab=\dfrac{1}{2}\) ta có \(\left\{{}\begin{matrix}ab=\dfrac{1}{2}\\a^2+b^2=1\end{matrix}\right.\)

\(\sqrt{3x^2+6x+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}\)
\(\Leftrightarrow\sqrt{3\left(x^2+2x\right)+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}\)
Đặt \(t=x^2+2x\), \(\left(t\ge0\right)\) phương trình trở thành:
\(\sqrt{3t+16}+\sqrt{t}=2\sqrt{t+4}\)
\(3t+16+t+2\sqrt{t\left(3t+16\right)}=4\left(t+4\right)\)
\(\Leftrightarrow2\sqrt{t\left(3t+16\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\3t+16=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\) \(\Leftrightarrow t=0\).
Với \(t=0\Leftrightarrow x^2+2x=0\) \(\Leftrightarrow x\left(x+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\).
Vậy x = 0 hoặc x = -2.

Ta có \(\dfrac{1}{\sqrt{3x+1}}=\dfrac{f'\left(x\right)}{f\left(x\right)}\)

\(\Rightarrow\int\dfrac{1}{\sqrt{3x+1}}dx=\int\dfrac{f'\left(x\right)}{f\left(x\right)}dx\)

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

\(\Rightarrow\dfrac{2}{3}.\sqrt{3x+1}+C=\ln\left|f\left(x\right)\right|=\ln\left|f\left(x\right)\right|\)

\(\Rightarrow f\left(x\right)=e^{\dfrac{2}{3}.\sqrt{3x+1}+C}\)

Mặt khác ta có f(1) = \(e^{\dfrac{4}{3}+C}=1\Rightarrow C=-\dfrac{4}{3}\)

Vậy nên f(x) = \(e^{\dfrac{2}{3}.\sqrt{3x+1}-\dfrac{4}{3}}\)

Từ đó ta tính được f(5) = \(e^{\dfrac{4}{3}}\)