Bài 2:

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

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

\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)

\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)

Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)

Bài 1:

a)x²-4x²y+4xy

=x(x-4xy+y)

b)đề sai

Bài 3:

3yx + 6x - y = 7

<=> x(3y+6) - (3y+6) = 27

<=> (3y+6)(x+1) = 27

Ta có bảng sau:

Vậy...

a) \(A=-4x^2-8x+3=-4\left(x^2+2x+1\right)+7=-4\left(x+1\right)^2+7\le7\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy Max(A) = 7 khi x = -1

b) \(B=6x-x^2+2=-\left(x^2-6x+9\right)+11=-\left(x-3\right)^2+11\le11\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)

Vậy Max(B) = 11 khi x = 3

c) \(C=x\left(2-3x\right)=-3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{1}{3}=-3\left(x-\frac{1}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\left(\forall x\right)\)

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

Vậy Max(C) = 1/3 khi x = 1/3

d) \(D=3x-x^2+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

Vậy Max(D) = 17/4 khi x = 3/2

e) \(E=3-2x^2+2xy-y^2-2x\)

\(E=-\left(x^2-2xy+y^2\right)-\left(x^2+2x+1\right)+4\)

\(E=-\left(x-y\right)^2-\left(x+1\right)^2+4\le4\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\Rightarrow x=y=-1\)

Vậy Max(E) = 4 khi x = y = -1

A = \(4x^2\) - 8x + 3

= [\(\left(2x\right)^2\) - 2.2x.2 + \(2^2\)] \(-2^2\) + 3

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

Ta có: \(\left(2x-2\right)^2\) ≤ 0 ∀ x

\(\left(2x-2\right)^2\) - 1 ≤ - 1

Hay A ≤ - 1

Dấu "=" xảy ra ↔ 2x - 2 = 0

2x = 2

x = 1

Vậy GTLN của A = - 1 ↔ x = 1

B = 6x \(-x^2\) + 2

= - (\(x^2\) - 6x) + 2

= - (\(x^2\) - 2.x.3 + \(3^2\)) \(-3^2\) + 2

= - \(\left(x-3\right)^2\) -7

Ta có: \(-\left(x-3\right)^2\) ≤ 0 ∀ x

\(-\left(x-3\right)^2\) - 7 ≤ - 7

Hay B ≤ - 7

Dấu "=" xảy ra ↔ - (x - 3) = 0

- x + 3 = 0

- x= - 3

x = 3

Vậy GTLN của B = - 7 ↔ x = 3

C = x(2 - 3x)

= 2x \(-3x^2\)

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

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

Ta có: \(-3\left(x+\frac{3}{4}\right)^2\) ≤ 0 ∀ x

\(-3\left(x+\frac{3}{4}\right)^2\) \(-\frac{9}{16}\) ≤ \(-\frac{9}{16}\)

Hay C ≤ \(-\frac{9}{16}\)

Dấu "=" xảy ra ↔ \(-3\left(x+\frac{3}{4}\right)\) = 0

- 3x \(-\frac{9}{4}\) = 0

- 3x = \(\frac{9}{4}\)

x = \(-\frac{3}{4}\)

Vậy GTLN của C = \(-\frac{9}{16}\) ↔ x = \(-\frac{3}{4}\)

\(A=\dfrac{3x^2+9x+17}{3x^2+9x+7}=1+\dfrac{10}{3x^2+9x+7}=1+\dfrac{10}{3\left(x^2+2.x.\dfrac{9}{2}+\dfrac{81}{4}\right)-\dfrac{215}{4}}\\ =1+\dfrac{10}{3\left(x+\dfrac{9}{2}\right)^2-\dfrac{215}{4}}\le\dfrac{35}{43}\)

Câu khác giải TT

1. Thu gọn biểu thức - Hoc24 làm rồi mà bạn?

1.

a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)

b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)

2.

a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)

b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

1,a,=x2−6x+8+3x2−15x=4x2−21x+8b,=9x2+12x+4−x2+9=8x2+12x+132,a,⇔x2+8x+16−x2+4=5⇔8x=−15⇔x=−158b,⇔9x2−6x+1−8x2−2x+12x+3−x2=5⇔4x=1⇔x=14

`a)|2x+1|=5`

`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

`b)|2x+1|=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

`c)|2x+1|=7`

`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\) 

`d)|2x+5|=|3x-7|`

`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\) 

`e)|2x+7|=1`

`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\) 

`g)|x-2|+|2x-3|=2`

Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`

`pt<=>x-2+2x-3=2`

`<=>3x-5=2`

`<=>3x=7`

`<=>x=7/3(tm)`

Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`

`pt<=>2-x+3-2x=2`

`<=>5-3x=2`

`<=>3x=3`

`<=>x=1(tm)`

Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`

`pt<=>2-x+2x-3=2`

`<=>x-1=2`

`<=>x=3(l)`

`h)|x+2|+|1-x|=3x+2`

Vì `VT>=0=>3x+2>=0=>x>=-2/3`

`=>|x+2|=x+2`

`pt<=>x+2+|1-x|=3x+2`

`<=>|1-x|=2x(x>=0)`

`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\) 

a.

$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=5\\ 2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

b.

$|2x+1|=0$

$\Leftrightarrow 2x+1=0$

$\Leftrightarrow x=-\frac{1}{2}$
c.

$|2x+1|=7$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)

d.

$|2x+5|=|3x-7|$

\(\Leftrightarrow \left[\begin{matrix} 2x+5=3x-7\\ 2x+5=7-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=12\\ x=0,4\end{matrix}\right.\)

e.

$|2x+7|=x-1\Rightarrow x-1\geq 0\Leftrightarrow x\geq 1$
Với $x\geq 1$ thì $|2x+7|=2x+7$

Khi đó pt trở thành:
$2x+7=x-1$

$\Leftrightarrow x=-8< 1$ (vô lý)

Vậy pt vô nghiệm.

g.

$|x-2|+|2x-3|=2$

Nếu $x\geq 2$ thì pt trở thành:

$x-2+2x-3=2$

$\Leftrightarrow 3x-5=2$

$\Leftrightarrow x=\frac{7}{3}$ (thỏa mãn)

Nếu $\frac{3}{2}\leq x< 2$ thì pt trở thành:

$2-x+2x-3=2$

$\Leftrightarrow x=3$ (không thỏa mãn)

Nếu $x< \frac{3}{2}$ thì pt trở thành:

$2-x+3-2x=2$

$\Leftrightarrow 5-3x=2$

$\Leftrightarrow x=1$ (thỏa mãn)

Vậy..........

h.

Từ đề suy ra $x\geq \frac{-2}{3}$

$\Rightarrow |x+2|=x+2$

Nếu  $x\geq 1$ thì $|1-x|=x-1$. PT trở thành:

$x+2+x-1=3x+2$

$\Leftrightarrow 2x+1=3x+2$

$\Leftrightarrow x=-1$ (vô lý)

Nếu $\frac{-2}{3}\leq x< 1$ thì $|1-x|=1-x$. PT trở thành:
$x+2+1-x=3x+2$

$\Leftrightarrow 3=3x+2$

$\Leftrightarrow x=\frac{1}{3}$ (thỏa mãn)

Lời giải:

a.

$|3x+1|=5$

$\Leftrightarrow 3x+1=\pm 5$

$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=-2$

b.

$2|2x-3|=\frac{2}{5}$

$\Leftrightarrow |2x-3|=\frac{1}{5}$

$\Leftrightarrow 2x-3=\pm \frac{1}{5}$

$\Leftrightarrow x=\frac{8}{5}$ hoặc $x=\frac{7}{5}$

c.

$|2-3x|=|5-2x|$

$\Leftrightarrow 2-3x=5-2x$ hoặc $2-3x=2x-5$

$\Leftrightarrow x=-3$ hoặc $x=1,4$

\(a,\left|3x+1\right|=5\)

\(\left|3x+1\right|=\left\{{}\begin{matrix}3x+1khix\ge-\dfrac{1}{3}\\-3x-1khix< -\dfrac{1}{3}\end{matrix}\right.\)

Với \(x\ge-\dfrac{1}{3}\Rightarrow3x+1=5\Rightarrow3x=4\Rightarrow x=\dfrac{4}{3}\left(tm\right)\)

Với \(x< -\dfrac{1}{3}\Rightarrow-3x-1=5\Rightarrow-3x=6\Rightarrow x=-2\left(tm\right)\)

Vậy \(S=\left\{-2;\dfrac{4}{3}\right\}\)

\(b,2\left|2x-3\right|=\dfrac{2}{5}\)

\(\Leftrightarrow\left|2x-3\right|=\dfrac{1}{5}\)

\(\left|2x-3\right|=\left\{{}\begin{matrix}2x-3khix\ge\dfrac{3}{2}\\-2x+3khix< \dfrac{3}{2}\end{matrix}\right.\)
Với \(x\ge\dfrac{3}{2}\Rightarrow2x-3=\dfrac{1}{5}\Rightarrow2x=\dfrac{16}{5}\Rightarrow x=\dfrac{8}{5}\left(tm\right)\)

Với \(x< \dfrac{3}{2}\Rightarrow-2x+3=\dfrac{1}{5}\Rightarrow-2x=-\dfrac{14}{5}\Rightarrow x=\dfrac{7}{5}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{8}{5};\dfrac{7}{5}\right\}\)

\(c,\left|2-3x\right|=\left|5-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=5-2x\\2-3x=-5+2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=3\\-5x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{7}{5}\end{matrix}\right.\)

Vậy \(S=\left\{-3;\dfrac{7}{5}\right\}\)

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