với x+y=1, ta có \(A=2\left(x+y\right)\left(x^2-xy+y^2\right)-3x^2-3y^2+30\)

                                \(=2\left(x^2-xy+y^2\right)-3x^2-3y^2+30=-x^2-2xy-y^2+30\)

                                  \(=-\left(x+y\right)^2+30=-1+30=-29\)

cậu Áp dụng bđt cô si để chứng minh \(\left(x+y\right)^2\ge4xy\)

Áp dụng ta có \(\left[a+\left(b+c\right)\right]^2\ge4a\left(b+c\right)\)

=> \(1\ge4a\left(b+c\right)\)(1)

Áp dụng lần nữa ta có 

\(\left(b+c\right)^2\ge4bc\) (2 )

từ (1),(2), nhận 2 vế ta có 

\(\left(b+c\right)^2\ge16\left(b+c\right)abc\)

=> \(b+c\ge16abc\) (ĐPCM) 

dấu = tự tìm nhé

\(\sqrt{x}+\sqrt{x+1}=\frac{1}{\sqrt{x}}\left(1\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+\sqrt{x+1}\right)=\sqrt{x}.\frac{1}{\sqrt{x}}\)

\(\Leftrightarrow x+\sqrt{x\left(x+1\right)}=1\)

\(\Leftrightarrow\sqrt{x\left(x+1\right)}=1-x\)

\(\Leftrightarrow\hept{\begin{cases}1-x\ge0\\x\left(x+1\right)=\left(1-x\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le1\\x^2+x=1-2x+x^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le1\\x^2+x-1+2x-x^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le1\\3x=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le1\\x=\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow x=\frac{1}{3}\)

Nhớ tk cho mk nha ahihi<:

x=0,3333333333

Áp dụng bđt AM - GM ta có :

\(\sqrt{b-1}\le\frac{b-1+1}{2}=\frac{b}{2}\Rightarrow a\sqrt{b-1}\le\frac{ab}{2}\)

\(\sqrt{a-1}\le\frac{a-1+1}{2}=\frac{a}{2}\Rightarrow b\sqrt{a-1}\le\frac{ba}{2}\)

\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le\frac{ab}{2}+\frac{ab}{2}=ab\)(đpcm)

b2 dễ tự lm

b2 x2 là x mũ 2. y2 là y mũ 2 .

yx−y=x​2​​+2

yx−y−x​2​​−2=0

x=​−2​​−y+√​y​2​​−4y−8​​​​​,​−2​​−y−√​y​2​​−4y−8​​​​​

x=​−2​​−y+√​y​2​​−4y−8​​​​​,​−2​​−y−√​y​2​​−4y−8​​​​​

x=−​2​​−y+√​y​2​​−4y−8​​​​​,−​2​​−y−√​y​2​​−4y−8​​​​​

k sau giúp tiếp 

a) \(BĐT\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)

\(\Leftrightarrow\sqrt{\frac{c\left(a-c\right)}{ab}}+\sqrt{\frac{c\left(b-c\right)}{ab}}\le1\)

\(\Leftrightarrow\sqrt{\frac{c}{b}\left(1-\frac{c}{a}\right)}+\sqrt{\frac{c}{a}\left(1-\frac{c}{b}\right)}\le1\)

Áp dụng AM-GM:\(VT\le\frac{1}{2}\left(\frac{c}{b}+1-\frac{c}{a}+\frac{c}{a}+1-\frac{c}{b}\right)=1\left(đpcm\right)\)

Dấu = xảy ra khi (a+b).c=ab

b) \(2+b+c+2+b+c\ge2\sqrt{\left(b+1\right)\left(c+1\right)}+2+b+c=\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge4\left(1+a\right)\)

\(\Leftrightarrow b+c\ge2a\)

cau a) dung cosi

\(\sqrt{c\left(a-c\right)}\le\frac{a-c+c}{2}\) ap dung cosi cho hai so c va a-c

tuong tu voi cac so khac

\(BT\le\frac{a-c+c}{2}+\frac{b-c+c}{2}-\frac{a+b}{2}\)(bt la VT cua de)

=> DPCM

b)

dung cosi nhu cau a

lam nhanh luon

\(\sqrt{1+b}\ge\frac{b+1+1}{2}\)

tuong tu

\(BT\ge\frac{b+2}{2}+\frac{c+2}{2}\ge a+2\)

<=> b+c>=2a

\(B=\sqrt{\left(7x-\frac{11}{7}\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}+\sqrt{\left(7x+\frac{11}{7}\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}\)

\(B=\sqrt{\left(\frac{11}{7}-7x\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}+\sqrt{\left(7x+\frac{11}{7}\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}\)

dùng Bất đẳng thức Bunyakovsky

\(B\ge\sqrt{\left(\frac{22}{7}\right)^2+\left(\frac{16\sqrt{5}}{7}\right)^2}\)

\(B\ge6\)

dấu "=" khi x=0

√​(9−√​7​​​)(9+√​7​​​)​​​x

√​9​2​​−√​7​​​​2​​​​​x

√​81−√​7​​​​2​​​​​x

√​81−7​​​x

√​74​​​x

cái 72 là 7 mũ 2 nha 

đặt \(A=-x^2-x+5=-\left(x^2+x-5\right)=-\left[\left(x^2+x+\frac{1}{4}\right)-\frac{21}{4}\right]\)

         \(=-\left(x+\frac{1}{2}\right)^2+\frac{21}{4}\)

mà \(-\left(x+\frac{1}{2}\right)^2+\frac{21}{4}\le\frac{21}{4}\)

=> \(A\le\frac{21}{4}\)

dấu = xảy ra <=> x=-1/2

Ta có \(x^2+x\ge0\)

Nên   \(-x^2-x\le0\)

Do đó \(-x^2-x+5\le5\)

Vậy giá trị lớn nhất của biểu thức là 5