Ta có: x² + 2y² - 2xy + 2x - 4y + 2 = 0

<=> (x - y)² + 2(x - y) + 1 + y² - 2y + 1 = 0

<=> (x - y + 1)² + (y - 1)² = 0

<=> \(\hept{\begin{cases}x-y+1=0\\y-1=0\end{cases}}\) <=> \(\hept{\begin{cases}y=1\\x=y-1=1-1=0\end{cases}}\)

Vậy (x;y) = {(0; 1)}

\(9x^2+12x+16-18x-24+9=49\)

\(9x^2-6x+1=49\)

\(9x^2-6x-48=0\)

\(9x^2-24x+18x-48=0\)

\(9x\left(x+2\right)-24\left(x+2\right)=0\)

\(\left(x+2\right)\left(9x-24\right)=0\)

\(\orbr{\begin{cases}x+2=0\\9x-24=0\end{cases}\orbr{\begin{cases}x=-2\left(tm\right)\\x=\frac{24}{9}\left(tm\right)\end{cases}}}\)

Trả lời:

x² + 6x - y² + 9

= ( x² + 6x + 9 ) - y²

= ( x² + 2.x.3 + 3² ) - y²

= ( x + 3 )² - y² 

= ( x + 3 - y ) ( x + 3 + y )

\(x^2+6x-y^2+9\)

\(=\left(x^2+6x+9\right)-y^2\)

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

\(=\left(x+3-y\right)\left(x+3+y\right)\)

 Ta có : \(\Sigma\dfrac{ab}{a^2+b^2}=3-\Sigma\dfrac{a^2+b^2-ab}{a^2+b^2}\) 

Thấy : \(0< ab\left(a^2+b^2-ab\right)\le\dfrac{\left(a^2+b^2\right)^2}{4}\)   

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

CMTT ; ta có : \(\dfrac{b^2+c^2-bc}{b^2+c^2}\le\dfrac{1}{4}\left(\dfrac{b}{c}+\dfrac{c}{b}\right);\dfrac{c^2+a^2-ac}{a^2+c^2}\le\dfrac{1}{4}\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)

Suy ra : \(\Sigma\dfrac{ab}{a^2+b^2}\ge3-\dfrac{1}{4}\left(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}\right)=\dfrac{1}{4}\left(\dfrac{a+c}{b}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\right)\)

Thấy : \(\dfrac{a+c}{b}+\dfrac{b+c}{a}+\dfrac{a+b}{c}=\dfrac{\left(a+c\right)ac+\left(b+c\right)bc+ab\left(a+b\right)}{abc}=ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)( do abc = 1 ) 

Áp dụng BĐT Schur ta được : \(ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\le a^3+b^3+c^3+3abc=\Sigma a^3+3\)   

Suy ra : \(\Sigma\dfrac{ab}{a^2+b^2}\ge3-\dfrac{1}{4}\left(\Sigma a^3+3\right)=\dfrac{9}{4}-\dfrac{1}{4}\Sigma a^3\cdot\)

Khi đó : \(\Sigma a^3+\Sigma\dfrac{ab}{a^2+b^2}\ge\dfrac{3}{4}\Sigma a^3+\dfrac{9}{4}\ge\dfrac{3}{4}.3+\dfrac{9}{4}=\dfrac{9}{2}\)

" = " <=> a = b = c = 1 

Vậy ... 

Khuya rồi còn đăng à bạn ? 

Ta có :\(B=x^2+2xy+y^2+2x+2y+10\)

\(=\left(x+y\right)^2+2\left(x+y\right)+10\)

\(=\left(x+y+1\right)^2+9\ge9\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow x+y+1=0\)

\(\Leftrightarrow x+y=-1\)

Vậy \(MinB=9\Leftrightarrow x+y=-1\)

\(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)

\(=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2+4a^2-4ab+b^2\) 

\(=2\left(a+b\right)^2+2c^2+4a^2-4ab+b^2\)

\(=2a^2+2b^2+4ab+2c^2+4a^2-4ab+b^2\)

\(=6a^2+3b^2+2c^2\)

(*) \(\Leftrightarrow4sinx.cosx+1=sinx-cosx\)

Đặt a = sin x ; b = cos x \(\left(-1\le a;b\le1\right)\) . Ta có : 

\(\hept{\begin{cases}a^2+b^2=1\left(1\right)\\4ab+1=a-b\left(2\right)\end{cases}}\)

(2) <=> : \(a\left(4b-1\right)=-b-1\) 

TH 1 : \(b=\frac{1}{4}\) ko t/m

TH 2 : \(b\ne\frac{1}{4}\) ; ta có : \(a=\frac{b+1}{1-4b}\)

Thay vào (1) được : \(\left(\frac{b+1}{1-4b}\right)^2+b^2=1\Leftrightarrow\left(b+1\right)^2+b^2\left(1-4b\right)^2=\left(1-4b\right)^2\)

\(\Leftrightarrow b^2+2b+1+b^2\left(16b^2-8b+1\right)=16b^2-8b+1\)

\(\Leftrightarrow16b^4-8b^3+2b^2+2b+1=16b^2-8b+1\)

\(\Leftrightarrow16b^4-8b^3-14b^2+10b=0\)

\(\Leftrightarrow8b^4-4b^3-7b^2+5b=0\)

\(\Leftrightarrow b\left(8b^3-4b^2-7b+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=0\\b=-1\end{cases}}\)

Với b = 0 ; suy ra : a = 1 ( t/m )  Suy ra L \(\hept{\begin{cases}sinx=1\\cosx=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{\pi}{2}+2k\pi\\x=\frac{\pi}{2}+k\pi\end{cases}}\Leftrightarrow x=\frac{\pi}{2}+2k\pi}\) ( k thuộc Z )

Với b = - 1 ; suy ra a = 0 ; làm tương tự  

Ko chắc 

Đặt \(sinx-cosx=t,t\in\left[-\sqrt{2},\sqrt{2}\right]\).

\(\Rightarrow t^2=\left(sinx-cosx\right)^2=sin^2x+cos^2x-sin2x=1-sin2x\)

\(\Leftrightarrow sin2x=1-t^2\)

Phương trình ban đầu tương đương với: 

\(2\left(1-t^2\right)=t-1\)

\(\Leftrightarrow\orbr{\begin{cases}t=1\left(tm\right)\\t=-\frac{3}{2}\left(l\right)\end{cases}}\)

Với \(t=1\):

\(sinx-cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{cases}}\left(k\inℤ\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{cases}}\left(k\inℤ\right)\)

Đáp án:

Giải thích các bước giải:

 $=\frac{(5-3).(5+3).(5^2+3^2).(5^4+3^4)....(5^{64}+3^{64})}{2}+$ $\frac{5^{128}-3^{128}}{2}$

$=\frac{(5-3).(5+3).(5^2+3^2).(5^4+3^4)....(5^{64}+3^{64})+5^{128}-3^{128}}{2}$

$=\frac{(5^{64}-3^{64}).(5^{64}+3^{64})+5^{128}-3^{128}}{2}$

$=2.5^{128}2$

$=5^{128}$

(2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^4-1)(2^4+1)....(2^32+1)-2^64

=......

=(2^32-1)(2^32+1)-2^64

=2^64-1-2^64=-1