a/ \(u_6=u_1+5d=8\Rightarrow u_1=8-5d\)

\(u_2=u_1+d;u_4=u_1+3d\)

\(\Rightarrow\left\{{}\begin{matrix}u_2=8-5d+d=8-4d\\u_4=8-5d+3d=8-2d\end{matrix}\right.\)

\(\Rightarrow\left(8-4d\right)^2+\left(8-2d\right)^2=16\Rightarrow...\)

b/ Câu này làm theo ý hiểu thôi, ko chắc đâu

\(Xet-S_n:\)

\(u_1=u_1\)

\(u_2=u_1+d\)

\(u_3=u_1+2d\)

......

\(u_n=u_1+\left(n-1\right)d\)

\(\Rightarrow S_n=u_1+u_2+...+u_n=u_1+u_1+d+...+u_1+\left(n-1\right)d=n.u_1+d+2d+....+\left(n-1\right)d\)

\(=n.u_1+\left(1+2+...+\left(n-1\right)\right)d=n.u_1+\dfrac{d\left(n-1\right).n}{2}=\dfrac{n\left[2u_1+\left(n-1\right)d\right]}{2}\)

Tương tụ với S(2n)

\(S_{2n}=u_1+u_2+...+u_{2n}=u_1+u_1+d+....+u_1+\left(2n-1\right)d\)

\(=2n.u_1+d+2d+...+\left(2n-1\right)d=2n.u_1+\left(1+2+...+\left(2n-1\right)\right)d=2n.u_1+d.n\left(2n-2\right)=2n\left(u_1+\left(n-1\right).d\right)\)

\(4S_n=S_{2n}\Leftrightarrow4.\dfrac{n\left[2u_1+\left(n-1\right)d\right]}{2}=2n\left(u_1+\left(n-1\right).d\right)\)

\(\Leftrightarrow2n\left[2u_1+\left(n-1\right)d\right]=2n\left[u_1+\left(n-1\right)d\right]\)\(\Leftrightarrow2u_1=u_1\Rightarrow u_1=0\)

\(u_5=u_1+4d=18\Rightarrow d=\dfrac{18}{4}=4,5\)

Ok check lại số má hộ tui nhó

a.

\(\left\{{}\begin{matrix}u_1+\left(u_1+4d\right)-\left(u_1+2d\right)=10\\\left(u_1+d\right)+\left(u_1+4d\right)=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u_1+2d=10\\2u_1+5d=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=36\\d=-13\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}u_1+d+u_1+3d=5\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4d=5-2u_1\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)

\(\Rightarrow u_1^2+\left(u_1+5-2u_1\right)^2=25\)

\(\Rightarrow u_1^2+u_1^2-10u_1+25=25\)

\(\Rightarrow\left[{}\begin{matrix}u_1=0\Rightarrow d=\dfrac{5}{4}\\u_1=5\Rightarrow d=-\dfrac{5}{4}\end{matrix}\right.\)

a/

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+x+y=20\)

\(\Leftrightarrow\left(x+y\right)^2+x+y-20=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=3\\x+y=-5\Rightarrow xy=12\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) theo Viet đảo x; y là nghiệm:

\(t^2-4t+3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=12\end{matrix}\right.\) theo Viet đảo x; y là nghiệm:

\(t^2+5t+12=0\left(vn\right)\)

b/

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy+1=0\\\left(x+y\right)^2-2xy-x-y=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+2xy+2=0\\\left(x+y\right)^2-2xy-x-y-22=0\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+\left(x+y\right)-20=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=-5\\x+y=-5\Rightarrow xy=4\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=-5\end{matrix}\right.\) thì x; y là nghiệm:

\(t^2-4t-5=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=5\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(-1;5\right);\left(5;-1\right)\)

TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=4\end{matrix}\right.\) thì x; y là nghiệm:

\(t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(-1;-4\right);\left(-4;-1\right)\)

c/

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+x=a\\y^2+y=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=8\\ab=12\end{matrix}\right.\) theo Viet đảo, a và b là nghiệm:

\(t^2-8t+12=0\Rightarrow\left[{}\begin{matrix}t=6\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=2\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x-6=0\\y^2+y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x-2=0\\y^2+y-6=0\end{matrix}\right.\end{matrix}\right.\)

Bạn tự bấm máy

Ta có: 

\(4\le\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=\sqrt{ab}+\sqrt{a}+\sqrt{b}+1\le\dfrac{a+b}{2}+\dfrac{a+1}{2}+\dfrac{b+1}{2}+1\)

\(=a+b+2\)

\(\Leftrightarrow a+b\ge2\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge\dfrac{\left(a+b\right)^2}{a+b}=a+b\ge2\)

Dấu \(=\) xảy ra khi \(a=b=1\).

2: Điểm rơi... đẹp!

Áp dụng bất đẳng thức AM - GM:

\(\left\{{}\begin{matrix}a^2+1\ge2a\\b^2+4\ge4b\\c^2+9\ge6c\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2+14\ge2\left(a+2b+3c\right)=28\).

\(\Rightarrow a^2+b^2+c^2\ge14\).

Đẳng thức xảy ra khi a = 1; b = 2; c = 3.

1: Ta có \(y^2\ge6-x+x-2=4\Rightarrow y\ge2\). 

Đẳng thức xảy ra khi x = 6 hoặc x = 2

\(y^2\le2\left(6-x+x-2\right)=8\Rightarrow y\le2\sqrt{2}\).

Đẳng thức xảy ra khi x = 4.

Cộng vế với vế:

\(x^2+2xy+y^2+x+y=12\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)-12=0\)

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

TH1: \(\left\{{}\begin{matrix}x+y=-4\\xy=5-\left(x+y\right)=9\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm: \(t^2-4t+9=0\) (vô nghiệm)

TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=5-\left(x+y\right)=2\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm:

\(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)