B = 2\(x^2\) - 4\(x\) - 8

B = 2(\(x^2\) - 2\(x\) + 4)  - 16

B = 2(\(x-2\))² - 16 

Vì (\(x-2\))² ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))² ≥ 0 ∀ \(x\)

⇒ 2(\(x-2\))²  - 16 ≥ -16 ∀ \(x\)

Dấu bằng xảy ra khi  (\(x-2\))² = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)

Vậy B_min = -16 khi \(x=2\)

Tìm min của C biết:

C = \(x^2\) - 2\(xy\) + 2y² + 2\(x\) - 10y + 17

C = (\(x^2\) - 2\(xy\) + y²) + 2(\(x\) - y) + y² - 8y + 16 + 1

C = (\(x\) - y)² + 2(\(x\) - y) + 1  + (y² - 8y + 16) 

C = (\(x-y+1\))² + (y - 4)² 

Vì (\(x\) - y + 1)² ≥ 0 ∀ \(x;y\); (y - 4)² ≥ 0 ∀ y

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy C_min = 0 khi (\(x;y\)) = (3; 4)

D = \(x^2\) - \(xy\) + y² - 2\(x\) - 2y

D=[\(x^2\)-2\(x\)\(\dfrac{y}{2}\)+(\(\dfrac{y}{2}\))²]-(2\(x\)-2\(\dfrac{y}{2}\)) +1 +(\(\dfrac{3}{4}\)y²-2.\(\dfrac{\sqrt{3}}{2}\)y .\(\sqrt{3}\) +3) - 4

D = (\(x-\dfrac{y}{2}\))² - 2(\(x-\dfrac{y}{2}\))+ 1 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

D = (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

Vì (\(x-\dfrac{y}{2}\) - 1)² ≥  0 ∀ \(x\);y và (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² ≥ 0 ∀ y 

Vậy (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4 ≥ - 4 ∀ \(x;y\)

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\dfrac{\sqrt{3}}{2}y-\sqrt{3}=0\end{matrix}\right.\)

      ⇒ \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\sqrt{3}.\left(\dfrac{1}{2}y-1\right)=0\end{matrix}\right.\)

  ⇒ \(\left\{{}\begin{matrix}x=1+\dfrac{1}{2}y\\\dfrac{1}{2}y=1\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=1+1\\y=1:\dfrac{1}{2}\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy D_min = - 4 khi (\(x;y\)) =(2; 2)

4, x^2-10x+8x-80=0

x(x-8)+10(x-8)=0

x+10=0         =)x=-10

hoặc

x-8=0        =)x=8

1, =(x+2)(x-2)=0

x+2=0     =)x=-2

hoặc

x-2=0      =)x=2

2,3(x^2-5^2)=0

=x+5=0    =)x=-5

hoặc 

 x-5=0   =)x=5

3,(3+2)^2=25

5^2=25

5, x^2-x-11x+11=0

x(x-1)-11(x-1)=0

x-11=0    =)x=11

hoặc 

x-1=0    =)x=1

xl nheee mk làm nhầm câu 4 trc

là mũ hay nhân vậy ?

bn có thể viết đề rõ hơn k nhìn có vẻ rối

Lời giải:

1.

$4x+9=0$

$4x=-9$

$x=\frac{-9}{4}$
2.

$-5x+6=0$

$-5x=-6$

$x=\frac{6}{5}$

3.

$x^2-1=0$

$x^2=1=1^2=(-1)^2$

$x=\pm 1$

4.

$x^2-9=0$

$x^2=9=3^2=(-3)^2$

$x=\pm 3$

5.

$x^2-x=0$

$x(x-1)=0$

$x=0$ hoặc $x-1=0$

$x=0$ hoặc $x=1$

6.

$x^2-2x=0$

$x(x-2)=0$

$x=0$ hoặc $x-2=0$

$x=0$ hoặc $x=2$

7.

$x^2-3x=0$

$x(x-3)=0$

$x=0$ hoặc $x-3=0$ 

$x=0$ hoặc $x=3$

8.

$3x^2-4x=0$

$x(3x-4)=0$

$x=0$ hoặc $3x-4=0$

$x=0$ hoặc $x=\frac{4}{3}$

\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)

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

\(=2\left[\left(x-1\right)^2-5\right]\)

\(=2\left(x-1\right)^2-10\ge-10\)

Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x+4=t\)

\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1\ge-1\)

hay \(\left(x^2+5x+5\right)^2-1\ge-1\)

Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)

\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)

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

\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)

Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)

\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)

\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)

\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)

Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)

18 - 4\(x\) = -20 - 6\(x\) 

-4\(x\) + 6\(x\) = - 20  - 18

      2\(x\)    = - 38

         \(x\)  = - 19

h, -15 \(\times\) 24 = -7\(x\) + 32

     7\(x\) = 360 + 32

      7\(x\) = 392

         \(x\) =  392:7

          \(x\) = 56

i, 15\(x\) -3.(4\(x\) - 6) = -12 + 36

    15\(x\) - 12\(x\) + 18 =  24

      3\(x\) = 24 - 18

       3\(x\) = 6

         \(x\) = 2

k, -10\(x\) - 27 = -7\(x\) + 33

    -27 - 33 = -7\(x\) + 10\(x\)

     3\(x\) = -60

        \(x\) = -20

m, -17\(x\) - 24 = - 9\(x\) - 40

     - 24 + 40 =  -9\(x\) + 17\(x\)

          8\(x\) = 16

             \(x\) = 2

n, -23 \(\times\) 25 = -18\(x\) + 75

     -575 = -18\(x\) + 75

       18\(x\) = 575 + 75

        18\(x\) = 650

              \(x\) = \(\dfrac{325}{9}\)

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

(-75) : (x + 6) = -15

x + 6 = -75 : (-15)

x + 6 = 5

x = 5 - 6

x = -1

B) -25 - 5.(x + 2) = 50

5(x + 2) = -25 - 50

5(x + 2) = -75

x + 2 = -75 : 5

x + 2 = - 15

x = -15 - 2

x = -17

D) 6.(2x - 8) - 8.(3x - 5) = -56

12x - 24 - 24x + 40 = -56

-12x + 16 = -56

-12x = -56 - 16

-12x = -72

x = (-72) : (-12)

x = 6

Lời giải:

a. 

$-75:(x+6)=-15$

$x+6=(-75):(-15)=5$

$x=5-6=-1$

b. 

$-25-5(x+2)=50$

$5(x+2)=-25-50=-75$

$x+2=-75:5=-15$

$x=-15-2=-17$

d.

$6(2x-8)-8(3x-5)=-56$

$(12x-48)-(24x-40)=-56$

$-12x-8=-56$

$-12x=-56+8=-48$

$x=(-48):(-12)=4$

(Đề ghi không rõ lắm, nên câu b có thể làm sai đề)

\(\dfrac{-75}{x+6}=-15\)

\(x+6=\dfrac{-75}{-15}=5\)

\(x=5-6=-1\)

\(-25-5x\left(x+2\right)=50\)

\(5x\left(x+2\right)=-25-50=-75\)

\(x\left(x+2\right)=x^2+2x=\dfrac{-75}{5}=-15\)

\(x^2+2x+1=-15+1=-14\)

\(\left(x+1\right)^2=-14\)

Suy ra không có nghiệm, vì không có căn bậc hai của số âm