a.

Tìm min:

$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$

Vậy $y_{\min}=2$

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Mặt khác: 

$y=4\sin x(\sin x+1)-8(\sin x+1)+11$

$=(\sin x+1)(4\sin x-8)+11$

$=4(\sin x+1)(\sin x-2)+11$

Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$

$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$

$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$

Vậy $y_{\max}=11$

b.

$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$

$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$

Vậy $y_{\max}=4$.

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Mặt khác:

$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$

$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$

$=(1+\sin x)(3-\sin x)$

Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$

$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$

Vậy $y_{\min}=0$

c.

$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$

$=\sin ^4x+2\sin ^2x-1$

$=(\sin ^4x-1)+(2\sin ^2x-2)+2$

$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$

$=(\sin ^2x-1)(\sin ^2x+3)+2$

Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$

$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$

$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$

Vậy $y_{\max}=2$

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$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$

Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$

$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$

\(a,A=-x^2-4x-2\)

\(=-\left(x^2+4x+2\right)\)

\(=-\left(x^2+4x+4\right)+2\)

\(=-\left(x^2+2\cdot x\cdot2+2^2\right)+2\)

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

Ta thấy: \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x+2\right)^2\le0\forall x\)

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

Dấu \("="\) xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy \(Max_A=2\) khi \(x=-2\).

Cậu xem lại giúp mình có sai đề bài không nhé!

#\(Toru\)

\(A=5\left(x^2-\dfrac{1}{5}x+\dfrac{1}{100}\right)+\dfrac{39}{20}=5\left(x-\dfrac{1}{10}\right)^2+\dfrac{39}{20}\ge\dfrac{39}{20}\)

\(A_{min}=\dfrac{39}{20}\) khi \(x=\dfrac{1}{10}\)

\(B=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+2\left(y^2-\dfrac{1}{2}y+\dfrac{1}{16}\right)-\dfrac{269}{24}=3\left(x+\dfrac{1}{6}\right)^2+2\left(y-\dfrac{1}{4}\right)^2-\dfrac{269}{24}\ge-\dfrac{269}{24}\)

\(B_{min}=-\dfrac{269}{24}\) khi \(x=-\dfrac{1}{6};y=\dfrac{1}{4}\)

A= 5x²-xz+2

A= (√5.x)²-2.√5.x.\(\dfrac{\text{√5}}{10}\)+\(\dfrac{1}{20}+\dfrac{39}{20}\)

A=(√5.x-\(\dfrac{\text{√5}}{10}\))²+\(\dfrac{39}{20}\)≥\(\dfrac{39}{20}\)

Dấu "=" xảy ra ⇔ (√5.x-\(\dfrac{\text{√5}}{10}\))=0

⇔ √5.x=\(\dfrac{\text{√5}}{10}\) ⇔ x=\(\dfrac{1}{10}\)

Vậy GTNN của A=\(\dfrac{39}{20}\) tại x=\(\dfrac{1}{10}\)

a: Ta có: \(3\left|2x+5\right|\ge0\forall x\)

\(\Leftrightarrow3\left|2x+5\right|-7\ge-7\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)

c: ta có: \(\left(2x-3\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x-3\right)^2-14\ge-14\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

\(C=\left(x^2+\dfrac{y^2}{4}+4-xy+4x-2y\right)+\dfrac{3}{4}\left(y^2-4y+4\right)+1011\)

\(=\left(x-\dfrac{y}{2}+2\right)^2+\dfrac{3}{4}\left(y-2\right)^2+1011\ge1011\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-1;2\right)\)

a) Ta có: \(B=x^2+4y^2+4x-4y\)

\(=\left(x^2+4x+4\right)+\left(4y^2-4y+1\right)-5\)

\(=\left(x+2\right)^2+\left(2y-1\right)^2-5\ge-5\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-2;\dfrac{1}{2}\right)\)

a: -1<=cos2x<=1

=>3>=-3cos2x>=-3

=>7>=-3cos2x+4>=1

=>7>=y>=1

\(y_{min}=1\) khi \(cos2x=1\)

=>2x=k2pi

=>x=kpi

\(y_{max}=-1\) khi cos2x=-1

=>2x=pi+k2pi

=>x=pi/2+kpi

b: \(0< =sin^2x< =1\)

=>\(3< =sin^2x+3< =4\)

=>3<=y<=4

y min=3 khi sin^2x=0

=>sinx=0

=>x=kpi

y max=4 khi sin^2x=1

=>cos^2x=0

=>x=pi/2+kpi

c: \(y=sin2x+3\)

-1<=sin2x<=1

=>-1+3<=sin2x+3<=1+3

=>2<=y<=4

\(y_{min}=2\) khi sin 2x=-1

=>2x=-pi/2+k2pi

=>x=-pi/4+kpi

y max=4 khi sin2x=1

=>2x=pi/2+k2pi

=>x=pi/4+kpi

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)