Đặt \(A=7x^2+5x+3\)

\(A=\left(7x^2+5x+\frac{25}{28}\right)+\frac{59}{28}\)

\(A=7\left(x^2+\frac{5}{7}x+\frac{25}{196}\right)+\frac{59}{28}\)

\(A=7\left(x+\frac{5}{14}\right)^2+\frac{59}{28}\ge\frac{59}{28}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(7\left(x+\frac{5}{14}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-5}{14}\)

Vậy GTNN của \(A\) là \(\frac{59}{28}\) khi \(x=\frac{-5}{14}\)

Đặt \(B=-3x^2-3x+5\)

\(B=\left(-3x^2-3x-\frac{3}{4}\right)+\frac{23}{4}\)

\(B=-3\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}\)

\(B=-3\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\le\frac{23}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-3\left(x+\frac{1}{2}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-1}{2}\)

Vậy GTLN của \(B\) là \(\frac{23}{4}\) khi \(x=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

Ta có:

\(7x^2+5x+3=7\left(x^2+\frac{5}{7}x+\frac{3}{7}\right)\)

\(=7\left(x^2+\frac{5}{7}x+\frac{25}{196}+\frac{59}{196}\right)\)

\(=7\left(x+\frac{5}{14}\right)^2+\frac{59}{28}\ge\frac{59}{28}\)

\(-3x^2-3x+5=-3\left(x^2+x-5\right)\)

\(=-3\left(x^2+x+\frac{1}{4}-\frac{21}{4}\right)=-3\left(x+\frac{1}{2}\right)^2+\frac{63}{4}\le\frac{63}{4}\)

a) Ta có: \(7x^2+5x+3=7\left(x^2+\frac{5}{7}x\right)+3\)

\(=7\left(x^2+2.\frac{5}{14}x+\frac{25}{196}\right)+3=7\left(x+\frac{5}{14}\right)^2+\frac{59}{28}\ge\frac{59}{28}\)

Dấu "=" xảy ra \(\Leftrightarrow7\left(x+\frac{5}{14}\right)^2=0\Leftrightarrow x=\frac{-5}{14}\)

Vậy giá trị nhỏ nhất của biểu thức là ....

b) \(-3x^2-3x+5=-\left(3x^2+3x-\frac{3}{4}\right)+\frac{23}{4}\)

\(=-3\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}\)

\(=-3\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\le\frac{23}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy GTLN của biểu thức trên là ...

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

\(ĐKXĐ:\)\(\frac{5}{3}\le x\le\frac{7}{3}\)

\(A^2=3x-5+7-3x+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)

Áp dụng BĐT Cô-si ta có :

\(A^2\le2+\left(3x-5+7-3x\right)=4\)

Dấu  =  xảy ra \(\Leftrightarrow\)\(3x-5=7-3x\Leftrightarrow x=2\)

Vậy Max \(A^2=4\)suy ra Max A = 2 khi x = 2 

sửa lại đề 

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

thông cảm nha

A = 3x² - 7x + 8 = 3(x² - 7/3 + 49/36) + 47/12 = 3(x - 7/6)² + 47/12

Ta luôn có: 3(x - 7/6)² \(\ge\)0 \(\forall\)x

=> 3(x - 7/6)² + 47/12 \(\ge\)47/12 \(\forall\)x

Dấu "=" xảy ra khi: x - 7/6 = 0 <=> x = 7/6

Vậy Min của A = 47/12 tại x = 7/6

1/ 0, 71

2/ Tương tự 2 câu 1, 3 nhé!

3/ 11,25

Tick đúng nha! Thanks!

1) \(B=-7x^2+9\)

Do \(x^2\ge0\forall x\Rightarrow-7x^2\le0\forall x\)

\(\Rightarrow B=-7x^2+9\le9\)

\(maxB=9\Leftrightarrow x=0\)

2) \(C=2-\left(3x-4\right)^4\)

Do \(\left(3x-4\right)^4\ge0\forall x\Rightarrow-\left(3x-4\right)^4\le0\forall x\)

\(\Rightarrow C=2-\left(3x-4\right)^4\le2\)

\(maxC=2\Leftrightarrow x=\dfrac{4}{3}\)

3) \(D=\dfrac{1}{2}x^2+3\)

Do \(\dfrac{1}{2}x^2\ge0\forall x\Rightarrow D=\dfrac{1}{2}x^2+3\ge3\)

\(minD=3\Leftrightarrow x=0\)

4) \(E=\dfrac{2016}{2-x^2+3}=\dfrac{2016}{-x^2+5}\)

Do \(x^2\ge0\forall x\Rightarrow-x^2+5\le5\forall x\)

\(\Rightarrow E=\dfrac{2016}{-x^2+5}\ge\dfrac{2016}{5}\)

\(minE=\dfrac{2016}{5}\Leftrightarrow x=0\)

\(B=-7x^2+9\)

Vì \(-7x^2\le0\forall x\)

\(\Rightarrow-7x^2+9\le9\forall x\)

\(\Rightarrow B_{max}=9\Leftrightarrow-7x^2=0\Leftrightarrow x=0\)

\(C=2-\left(3x-4\right)^4\)

Vì \(-\left(3x-4\right)^4\le0\forall x\)

\(\Rightarrow-\left(3x-4\right)^4+2\le2\forall x\)

\(\Rightarrow C_{max}=2\Leftrightarrow-\left(3x-4\right)^4=0\Leftrightarrow x=\dfrac{4}{3}\)

Nếu tìm GTLN thì câu \(d\) là \(D=-\dfrac{1}{2}x^2+3\)

Vì \(-\dfrac{1}{2}x^2\le0\forall x\)

\(\Rightarrow-\dfrac{1}{2}x^2+3\le3\forall x\)

\(\Rightarrow D_{max}=3\Leftrightarrow-\dfrac{1}{2}x^2=0\Leftrightarrow x=0\)

\(E=\dfrac{2016}{2-x^2+3}=\dfrac{2016}{5-x^2}\)

Vì \(x^2\ge0\forall x\)

\(\Rightarrow5-x^2\le5\forall x\)

\(\Rightarrow E_{min}=5\Leftrightarrow x=\dfrac{2016}{5}\)

Answer:

3.

\(x^2+2y^2+2xy+7x+7y+10=0\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+7x+7y+y^2+10=0\)

\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)

\(\Rightarrow4S^2+28S+4y^2+40=0\)

\(\Rightarrow4S^2+28S+49+4y^2-9=0\)

\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)

\(\Rightarrow-3\le2S+7\le3\)

\(\Rightarrow-10\le2S\le-4\)

\(\Rightarrow-5\le S\le-2\left(2\right)\)

Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)

Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)

Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

a: \(B\left(x\right)=-\left(x^2-3x+7\right)\)

\(=-\left(x^2-3x+\dfrac{9}{4}+\dfrac{19}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}\)

Dấu '=' xảy ra khi x=3/2

b: Ta có: \(C\left(x\right)=-x^2+7x-20\)

\(=-\left(x^2-7x+20\right)\)

\(=-\left(x^2-7x+\dfrac{49}{4}+\dfrac{31}{4}\right)\)

\(=-\left(x-\dfrac{7}{2}\right)^2-\dfrac{31}{4}\le-\dfrac{31}{4}\)

Dấu '=' xảy ra khi x=7/2