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

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

1) x³ + 5x² + 3x - 9

= x³ + 2x² + 3x² + 6x - 3x - 9

= ( x³ + 2x² ) + (3x² + 6x ) - ( 3x + 9 )

= x² ( x+ 2 ) + 3x ( x + 2) - 3( x +2 )

= ( x + 2 ) ( x² + 3x -3 )

2) x³ + 5x² + 8x + 4

= ( x³ + x² ) + ( 4x² + 4x ) + ( 4x + 4 )

= x² ( x + 1 ) + 4x ( x + 1 ) + 4 ( x + 1 )

= ( x + 1) ( x² + 4x + 4 )

= (x + 1 ) ( x + 2 )²

3) x³ - 9x² + 6x + 16

= x³ - 8x² - x² + 8x - 2x + 16

= ( x³ - 8x² ) - ( x² - 8x ) - ( 2x - 16 )

= x² ( x - 8 ) - x ( x - 8 ) - 2 ( x - 8 )

= ( x - 8 ) ( x² - x - 2 )

4) x³ - 4x² + x + 6

= x³ - 3x² - x² + 3x - 2x + 6

= ( x³ - 3x² ) - ( x² - 3x ) - ( 2x - 6)

= x² ( x - 3 ) - x ( x- 3 ) - 2 ( x - 3)

= ( x - 3 ) ( x² - x - 2 )

ĐKXĐ: \(x\in R\)

\(3x^2-5x+6=2x\cdot\sqrt{x^2-x+2}\)

=>\(3x^2-6x+x-2+8=2\cdot\sqrt{x^4-x^3+2x^2}\)

=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\left(\sqrt{x^4-x^3+2x^2}-4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-x^3+2x^2-16}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-2x^3+x^3-2x^2+4x^2-8x+8x-16}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left(3x+1\right)=\dfrac{2\left(x-2\right)\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left[\left(3x+1\right)-\dfrac{2\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\right]=0\)

=>x-2=0

=>x=2(nhận)

\(3x^2-5x+6=2x\sqrt{x^2-x+2}\)

\(\Leftrightarrow\left[x^2-2x\sqrt{x^2-x+2}+\left(x^2-x+2\right)\right]+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x^2-x+2}\right)^2+\left(x-2\right)^2=0\)

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

Thử lại ta thấy nghiệm \(x=2\) thỏa phương trình ban đầu.

\(\Rightarrow\left(3x+2\right).\left(5x-3\right)=\left(5x+7\right)\left(3x-1\right)\)

\(\Rightarrow15x^2-9x+10x-6=15x^2-5x+21x-7\)

\(\Rightarrow19x-6=26x-7\)

\(\Rightarrow26x-19x=7-6\)

\(\Rightarrow13x=1\)

\(\Rightarrow x=\frac{1}{13}\)

\(a̸\)   

\(\frac{3}{5}-\frac{1}{2}.x=\frac{1}{4}\)

            \(\frac{1}{2}.x=\frac{3}{5}-\frac{1}{4}\)

           \(\frac{1}{2}.x=\frac{7}{20}\)

            \(\Rightarrow\frac{7}{10}\)

\(b̸\)

\(11,3+2\left[x-\frac{1}{3}\right]=\frac{25}{6}\)

\(2\left[x-\frac{1}{3}\right]=\frac{25}{6}-11,3\)

\(2\left[x-\frac{1}{3}\right]=\frac{-107}{15}\)

\(x-\frac{1}{3}=\frac{-107}{15}:2\)

 \(x-\frac{1}{3}=\frac{-107}{30}\)

\(x=\frac{-107}{30}+\frac{1}{3}\)

\(x=\frac{-97}{30}\)

\(a)\frac{3}{5}-\frac{1}{2}x=\frac{1}{4}\)

\(\implies\frac{1}{2}x=\frac{3}{5}-\frac{1}{4}\)

\(\implies\frac{1}{2}x=\frac{7}{20}\)

\(\implies x=\frac{7}{20}:\frac{1}{2}\)

\(\implies x=\frac{7}{10}\)

Vậy...

\(b) 11,3+2(x-\frac{1}{3})=\frac{25}{6}\)

\(\implies \frac{113}{10}+2x-2.\frac{1}{3}=\frac{25}{6}\)

\(\implies \frac{113}{10}+2x-\frac{2}{3}=\frac{25}{6}\)

\(\implies \frac{113}{10}+2x=\frac{25}{6}+\frac{2}{3}\)

\(\implies \frac{113}{10}+2x=\frac{29}{6}\)

\(\implies 2x=\frac{29}{6}-\frac{113}{10}\)

\(\implies 2x=\frac{-97}{15}\)

\(\implies x=\frac{-97}{30}\)

Vậy..

\(c)5x-435+2x+140+3x=565\)

\(\implies (5x+2x+3x)+(-435+140)=565\)

\(\implies 10x+(-295)=565\)

\(\implies 10x=565-(-295)\)

\(\implies 10x=860\)

\(\implies x=86\)

Vậy...

~ hok tốt a~