a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

`#3107.101107`

a)

`x^2 + 6x + 10`

`= (x^2 + 2*x*3 + 3^2) + 1`

`= (x + 3)^2 + 1`

Vì `(x + 3)^2 \ge 0` `AA` `x`

`=> (x + 3)^2 + 1 \ge 1` `AA` `x`

Vậy, GTNN của bt là 1 khi `(x + 3)^2 = 0`

`<=> x + 3 = 0`

`<=> x = -3`

b)

`4x^2 - 4x + 5`

`= [(2x)^2 - 2*2x*1 + 1^2] + 4`

`= (2x - 1)^2 + 4`

Vì `(2x - 1)^2 \ge 0` `AA` `x`

`=> (2x - 1)^2 + 4 \ge 4` `AA` `x`

Vậy, GTNN của bt là `4` khi `(2x - 1)^2 = 0`

`<=> 2x - 1 = 0`

`<=> 2x = 1`

`<=> x = 1/2`

c)

`x^2 - 3x + 1`

`= (x^2 - 2*x*3/2 + 9/4) - 5/4`

`= (x - 3/2)^2 - 5/4`

Vì `(x - 3/2)^2 \ge 0` `AA` `x`

`=> (x - 3/2)^2 - 5/4 \ge -5/4` `AA` `x`

Vậy, GTNN của bt là `-5/4` khi `(x - 3/2)^2 = 0`

`<=> x - 3/2 = 0`

`<=> x = 3/2`

A(x)+B(x)=-2x^4+x^3+x^2+5x-5-x^4-3x^3+4x^2-6x+7

=-3x^4+4x^3+5x^2-x+2

A(x)-B(x)=-2x^4+x^3+x^2+5x-5+x^4+3x^3-4x^2+6x-7

=-x^4+4x^3-3x^2+11x-2

B(x)-C(x)

=-x^4-3x^3+4x^2-6x+7-x^3-x+2

=-x^4-4x^3+4x^2-7x+9

Đáp án C

Điều kiện  4 x 2 − 4 x + 1 2 x > 0 ⇔ x > 0

P T ⇔ log 7 2 x − 1 2 + 2 x − 1 2 = 2 x + log 7 2 x ⇔ f 2 x − 1 2 = f 2 x

với  f t = log 7 t + t

f ' t = 1 t ln 7 + 1 > 0 với t >0

→ P T ⇔ 2 x = 2 x − 1 2 ⇔ x = 3 ± 5 4

⇒ x 1 + 2 x 2 = 9 ± 5 4 ⇒ a = 9 b = 5 ⇒ a + b = 14

Đáp án C

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)

a)4x²-9=0

⇔ (2x-3)(2x+3)=0

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

b)(x+5)²-(x-1)²=0

⇔ (x+5-x+1)(x+5+x-1)=0

⇔ 12(x+2)=0

⇔ x=-2

c)x²-6x-7=0

⇔ x²-7x+x-7=0

⇔ x(x-7)+(x-7)=0

⇔ (x-7)(x+1)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

d)(x+1)²-(2x-1)²=0

⇔ (x+1-2x+1)(x+1+2x-1)=0

⇔3x(2-x)=0

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

a, 4x² - 9 = 0

<=> 4x² = 9

<=> x² = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)

b, (x + 5 )² - ( x - 1 )² = 0

<=> ( x+5-x+1 )(x+5+x-1) = 0

<=> 6(2x+4) = 0

<=> 12x+24=0

<=> 12x = -24

<=> x = -2

c, x²-6x-7=0

<=> x²+x-7x-7=0

<=> x(x+1)-7(x+1)=0

<=> (x-7)(x+1)=0

=> x+7=0 hoặc x+1=0

+ x-7=0 => x=7

+ x+1=0 => x=-1

d, \(\left(x+1\right)^2-\left(2x-1\right)^2=0\)

<=> \(\left(x+1-2x+1\right)\left(x+1+2x-1\right)=0\)

<=> (-x+2).3x=0

=> x=0 hoặc (-x+2).3=0

+ (-x+2).3=0 => -3x+6=0 => x=-2

b) (x +5)² -(x -1)²=0

<=> [(x +5) -(x -1)][(x +5) +(x -1)]=0

<=> (x +5 -x +1)(x +5 +x -1)=0

<=> 6(2x+4)=0 <=>12(x +2)=0

=> x +2=0=> x=-2

vậy x= -2

c) x² -6x -7=0

<=> x² -7x +x -7=0

<=> (x² +x)( -7x -7)=0

<=> x(x +1).-7(x +1)=0

<=> (x +1)(x -7)=0

<=> \(\left\{{}\begin{matrix}x+1=0\\x-7=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)

Vậy S={-1; 7}

d) (x +1)² -(2x -1)²=0

<=> [(x -1)-(2x -1)][(x -1)+(2x -1)]=0

<=> (x -1 -2x +1)(x -1 +2x -1)=0

<=> (x -2x)(3x -2)<=> -x(3x -2)=0

<=> \(\left\{{}\begin{matrix}-x=0\\3x-2=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy S={0; \(\dfrac{2}{3}\)}