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 \(x=\dfrac{3}{2}\)

\(A=\dfrac{4\left(x^2-4x+4\right)+\left(x^2-8x+16\right)}{x^2-4x+4}=4+\left(\dfrac{x-4}{x-2}\right)^2\ge4\)

\(A_{min}=4\) khi \(x=4\) (A max ko tồn tại)

\(B=\dfrac{6\left(x^2+2x+1\right)+\left(4x^2+12x+9\right)}{x^2+2x+1}=6+\left(\dfrac{2x+3}{x+1}\right)^2\ge6\)

\(B_{min}=6\) khi \(x=-\dfrac{3}{2}\) 

B max ko tồn tại

bài 1:

a. \((x+1)(x+3) - x(x+2)=7 \)

    \(x^2+ 3x +x +3 - x^2 -2x =7\)

    \(x^2+4x+3-x^2-2x=7\)

\(=> 2x+3=7\)

    \(2x=4\)

    \(x = 2\)

Bài 2:

a)

\((3x-5)(2x+11) -(2x+3)(3x+7) \)

\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)

\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)

\(=-65\)

\(\)

a.

\(y=\dfrac{3}{2}sin2x-2\left(cos^2x-sin^2x\right)+5=\dfrac{3}{2}sin2x-2cos2x+5\)

\(=\dfrac{5}{2}\left(\dfrac{3}{5}sin2x-\dfrac{4}{5}cos2x\right)+5=\dfrac{5}{2}sin\left(2x-a\right)+5\) (với \(cosa=\dfrac{3}{5}\))

\(\Rightarrow-\dfrac{5}{2}+5\le y\le\dfrac{5}{2}+5\)

b.

\(\Leftrightarrow y.sinx-2y.cosx+4y=3sinx-cosx+1\)

\(\Leftrightarrow\left(y-3\right)sinx+\left(1-2y\right)cosx=1-4y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(y-3\right)^2+\left(1-2y\right)^2\ge\left(1-4y\right)^2\)

\(\Leftrightarrow11y^2+2y-9\le0\)

\(\Leftrightarrow-1\le y\le\dfrac{9}{11}\)

c.

Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)

\(\Rightarrow y=\dfrac{2\left(sin^2a+6sina.cosa\right)}{1+2sina.cosa+cos^2a}=\dfrac{1-cos2a+6sin2a}{1+sin2a+\dfrac{1+cos2a}{2}}=\dfrac{2-2cos2a+12sin2a}{3+2sin2a+cos2a}\)

\(\Leftrightarrow3y+2y.sin2a+y.cos2a=2-2cos2a+12sin2a\)

\(\Leftrightarrow\left(2y-12\right)sin2a+\left(y+2\right)cos2a=2-3y\)

Theo điều kiện có nghiệm của pt bậc nhất theo sin2a, cos2a:

\(\left(2y-12\right)^2+\left(y+2\right)^2\ge\left(2-3y\right)^2\)

\(\Leftrightarrow y^2+8y-36\le0\)

\(\Rightarrow-4-2\sqrt{13}\le y\le-4+2\sqrt{13}\)

#maianhhappy

bài 1 tính giá trị biểu thức

( - 25 ) nhân ( -3 ) nhân x với x = 4

\(\left(-25\right).\left(-3\right).4\)

\(=\left(-25\right).4.\left(-3\right)\)

\(=-100.\left(-3\right)=300\)

( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25

\(\left(-1\right).\left(-4\right).5.8.25\)

\(=4.5.8.25=4.25.5.8\)

\(=100.40=40000\)

( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12

\(\left(2.4.\left(-6\right)\right)^2:12\)

\(=\left(-48\right)^2:12\)

\(=2304:12=192\)

[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9

\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)

\(=-2700:\left(-9\right)\)

\(=300\)

(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3

\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)

\(=16:2.8\)

\(=8.8=64\)

bài 2 tìm x

( 2x _ 5 ) + 17 = 6

\(2x-5=-11\)

\(2x=-6\)

\(x=-3\)

10 _ 2 ( 4 _ 3x ) = - 4

\(2.\left(4-3x\right)=14\)

\(4-3x=7\)

\(3x=-3\)

\(x=-1\)

- 12 + 3 ( -x + 7 ) = -18

\(3\left(-x+7\right)=-6\)

\(-x+7=-2\)

\(-x=-9\)

\(x=9\)

24 : ( 3x _ 2 ) = -3

\(3x-2=-8\)

\(3x=-6\)

\(x=-2\)

- 45 : 5 nhân ( -3 _ 2x ) = 3

\(5.\left(-3-2x\right)=-15\)

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

\(2x=0\)

\(x=0\)

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 $y_{\min}=2$

----------------

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 $y_{\max}=4$.

---------------------------

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 $y_{\max}=2$

------------------------------------------

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

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 $y_{\min}=-1$