\(x^2+y^2-2x-4y-4=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2-9=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=9=0^2+3^2=0^2+\left(-3\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=3\\y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=0\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-3\\y-2=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-2\le x\le4\left(y\in R\right)\)

Ta có \(S=3x+4y\)

Mà \(x\ge-2;y\ge-1\Leftrightarrow S\ge3\cdot\left(-2\right)+4\cdot\left(-1\right)=-6-4=-10\)

Vậy GTNN của S là \(-10\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

Lời giải:

ĐKĐB $\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)-9=0$

$\Leftrightarrow (x-1)^2+(y-2)^2-9=0$

$\Rightarrow (x-1)^2=9-(y-2)^2\leq 9$

$\Rightarrow -3\leq x-1\leq 3$

$\Leftrightarrow -2\leq x\leq 4$

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Đặt $x-1=a; y-2=b$ thì bài toán trở thành:
Cho $a,b$ thực thỏa mãn $a^2+b^2=9$

Tìm min $S=3a+4b+11$

Áp dụng BĐT Bunhiacopxky:

$(3a+4b)^2\leq (a^2+b^2)(3^2+4^2)=9.25$

$\Rightarrow -15\leq 3a+4b\leq 15$

$\Rightarrow 3a+4b\geq -15$

$\Rightarrow S=3a+4b+11\geq -4$

Vậy $S_{\min}=-4$ khi $x=\frac{-4}{5}; y=\frac{-1}{5}$

a, \(\dfrac{4\left(x-3\right)^2-\left(2x-1\right)^2-12x}{12}< 0\)

\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+4x-1-12x< 0\)

\(\Leftrightarrow-32x+35< 0\Leftrightarrow x>\dfrac{35}{32}\)

b, \(\dfrac{24+12\left(x+1\right)-36+3\left(x-1\right)}{12}< 0\)

\(\Rightarrow-12x+15x+9< 0\Leftrightarrow3x< -9\Leftrightarrow x>-3\)

Lời giải:

Thay $x=0$ vào điều kiện đề thì $f(1)=0$ hoặc $f(1)=-1$ 

Đạo hàm 2 vế:

$4f(2x+1)f'(2x+1)_{2x+1}=1+3f(1-x)^2f'(1-x)_{1-x}$

Thay $x=0$ vô thì:

$4f(1)f'(1)=1+3f(1)^2f'(1)$

Nếu $f(1)=0$ thì hiển nhiên vô lý

Nếu $f(1)=-1$ thì: $-4f'(1)=1+3f'(1)\Rightarrow f'(1)=\frac{-1}{7}$

PTTT tại $x=1$ có dạng:

$y=f'(1)(x-1)+f(1)=\frac{-1}{7}(x-1)-1=\frac{-x}{7}-\frac{6}{7}$

\(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\Rightarrow f\left(2\right)=-1\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{f\left(x\right)+2x+1}+x}.\dfrac{\left(\sqrt{f\left(x\right)+2x+1}-x\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\dfrac{f\left(x\right)+1-x\left(x-2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\left(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}-\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{x-2}\right)\)

\(=\dfrac{1}{4\left(\sqrt{4}+2\right)}.\left(a-2\right)=\dfrac{a-2}{16}\)

1) 

(=)x² = 8² + 6² = 64+36=100=10² = (-10)² 

=> x=10 hoặc x=-10

2)

(=)|x-1| = -26/-24=13/12

=> x-1 = 13/12 hoặc x-1=-13/12

=> x= 25/12 hoặc x= -1/12

3) 

(2x-4+7)\(⋮\left(x-2\right)\) 

(=) 2(x-2) + 7 \(⋮\left(x-2\right)\)

(=) 7 \(⋮\left(x-2\right)\)

(=) x-2 \(\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

(=) x\(\in\left\{-5;1;3;9\right\}\)

vì x bé nhất => x=-5

#Học-tốt