=3(x^2-5/3x+1)

=3(x^2-2*x*5/6+25/36+11/36)

=3(x-5/6)^2+11/12>=11/12

Dấu = xảy ra khi x=5/6

Câu 1:

\(M=x^2-3x+5\)

\(M=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}\)

\(M=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

            Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

    Vậy Min M = 11/4 khi x=3/2

b)\(N=2x^2+3x\)

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

\(N=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)-\frac{9}{8}\)

\(N=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

              Dấu = xảy ra khi \(x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)

                       Vậy MIn N = -9/8 khi x=-3/4

c)Tự làm nha

Ta có : x² - 3x + 5 

= x² - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\) + \(\frac{11}{4}\)

= \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\in R\)

Nên : \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\) \(\ge\frac{11}{4}\forall x\in R\)

Vậy GTNN của biểu thức là : \(\frac{11}{4}\) khi \(x=\frac{3}{2}\)

Câu 2:

a)\(A=-x^2-5x+3\)

\(A=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{37}{4}\)

\(A=\frac{37}{4}-\left(x+\frac{5}{2}\right)^2\le\frac{37}{4}\)

            Dấu = xảy ra khi \(x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)

                      Vậy Max A = 37/4 khi x=-5/2

b)\(B=-2x^2+3x\)

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

\(B=-2\left(x^2-2.\frac{3}{4}+\frac{9}{16}\right)+\frac{9}{8}\)

\(B=\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\le\frac{9}{8}\)

         Dấu = xảy ra khi \(x-\frac{3}{4}=0\Rightarrow x=\frac{3}{4}\)

                    Vậy Max B=9/8 khi x=3/4

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 = 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

Bạn xem lại ĐKĐB. Nếu $x\geq \frac{-1}{3}$ thì mình nghi ngờ $\sqrt{3x-1}$ của bạn viết là $\sqrt{3x+1}$Còn nếu đúng là $\sqrt{3x-1}$ thì ĐK cần là $x\geq \frac{1}{3}$.

a)

\(f\left(x\right)=3x^2-5x+1\)

\(3f\left(x\right)=9x^2-15x+3\)

\(3f\left(x\right)=\left(9x^2-15x+\frac{25}{4}\right)-\frac{13}{4}\)

\(3f\left(x\right)=\left(3x-\frac{5}{2}\right)^2-\frac{13}{4}\)

Mà \(\left(3x-\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3f\left(x\right)\ge\frac{-13}{4}\)

\(\Leftrightarrow f\left(x\right)\ge-\frac{13}{12}\)

Dấu '=' xảy ra khi :

\(3x-\frac{5}{2}=0\Leftrightarrow3x=\frac{5}{2}\Leftrightarrow x=\frac{5}{6}\)

\(f\left(x\right)=2x^2-9x-3\)

\(2f\left(x\right)=4x^2-18x-6\)

\(2f\left(x\right)=\left(4x^2-18x+\frac{81}{4}\right)-\frac{105}{4}\)

\(2f\left(x\right)=\left(2x-\frac{9}{2}\right)^2-\frac{105}{4}\)

Mà \(\left(2x-\frac{9}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2f\left(x\right)\ge-\frac{105}{4}\)

\(\Leftrightarrow f\left(x\right)\ge-\frac{105}{8}\)

Dấu "=" xảy ra khi :

\(2x-\frac{9}{2}=0\Leftrightarrow2x=\frac{9}{2}\Leftrightarrow x=\frac{9}{4}\)

\(x^3+9x+26=0\)

\(\Leftrightarrow x^3+2x^2-2x^2-4x+13x+26=0\)

\(\Leftrightarrow x^2\left(x+2\right)-2x\left(x+2\right)+13\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+13\right)=0\) (1)

Ta có: \(x^2-2x+13=\left(x-1\right)^2+12\) >0 với mọi x

Khi đó: \(\left(1\right)\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy pt đã cho có nghiệm x=-2

\(I=3x^2+4xy+4y^2+5x=\left(2x^2+5x+\dfrac{25}{8}\right)+\left(x^2+4xy+4y^2\right)-\dfrac{25}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\left(x+2y\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\)

\(minI=-\dfrac{25}{8}\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-\dfrac{5}{4}\\y=\dfrac{5}{8}\end{matrix}\right.\)