9:(x+2)²+9=9,25

⇒9:(x+2)²+9=9,25

⇒9:(x+2)²+9=9,25

⇒9:(x+2)²=0,25

⇒(x+2)²=36

⇒hoặc x+2=√36⇒x+2=6⇒x=4

           x+2=-√36⇒x+2=-6⇒x=-8

vậy x={4;-8}

Ta có: \(\dfrac{9}{\left(x+2\right)^2}+9=9.25\)

\(\Leftrightarrow\dfrac{9}{\left(x+2\right)^2}=0.25\)

\(\Leftrightarrow\left(x+2\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

\(a,2^{x+1}=64\\ \Rightarrow a,2^{x+1}=2^6\\ \Rightarrow x+1=6\\ \Rightarrow x=5\)

\(b,x=18\)

\(c,\left(4x-9\right)-\left(x+111\right)=0\\ \Rightarrow4x-9-x-111=0\\ \Rightarrow3x-120=0\\ \Rightarrow3x=120\\ \Rightarrow x=40\)

a
,
2
x
+
1
=
64
⇒
a
,
2
x
+
1
=
2
6
⇒
x
+
1
=
6
⇒
x
=
5

b
,
x
=
18

c
,
(
4
x
−
9
)
−
(
x
+
111
)
=
0
⇒
4
x
−
9
−
x
−
111
=
0
⇒
3
x
−
120
=
0
⇒
3
x
=
120
⇒
x
=
40

Đề bài yêu cầu gì?

a) 2²(6x - 3²) - 3 = 33

4(6x - 9) = 33 + 3

4(6x - 9) = 36

6x - 9 = 36 : 4

6x - 9 = 9

6x = 9 + 9

6x = 18

x = 18 : 6

x = 3

b) 4(x + 2) = 3(x + 1) + 17

4x + 8 = 3x + 3 + 17

4x - 3x = 3 + 17 - 8

x = 12

Đề là gì vậy ???

\(B=x\left(x-2\right)\left(x+2\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(B=x\left(x^2-4\right)-\left(x^3-3x^2+9x+3x^2-9x+27\right)\)

\(B=x^3-4x-\left(x^3+27\right)\)

\(B=-4x-27\)

đề là rút gọn ạ

\(\dfrac{x}{15}\)+\(\dfrac{x}{12}\)=4/1+1/2=9/2

=>x(\(\dfrac{1}{15}\)+\(\dfrac{1}{12}\))=9/2

=>x\(\cdot\)\(\dfrac{3}{20}\)=9/2

=>x=9/2:3/20=30

Vậy x=30

\(\dfrac{x}{15}+\dfrac{x}{12}=\dfrac{9}{2}\Rightarrow\left(\dfrac{1}{15}+\dfrac{1}{12}\right)x=\dfrac{9}{2}\)

\(\Rightarrow\left(\dfrac{12+18}{180}\right)x=\dfrac{9}{2}\Rightarrow\dfrac{30}{180}x=\dfrac{9}{2}\Rightarrow\dfrac{1}{6}x=\dfrac{9}{2}\Rightarrow x=\dfrac{9}{2}.6=27\)

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`

`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`

`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`

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

`=(2x+6)/(x-2)`

ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)

\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)

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

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

\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)

\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)

Chắc mình làm tắt quá để mình làm lại bước biến đổi.

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((x-3)/(x+3)+(4x^2)/(x^2-9)-(x+3)/(x-3)):((2x+1-x-3)/(x+3))`

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

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

`=(x^2-6x+9+4x^2-x^2-6x-9)/(x^2-9)*(x+3)/(x-2)`

`=(4x^2-12x)/(x^2-9)*(x+3)/(x-2)`

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

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

`=(4x)/(x-2)`

k cho mik nhá

\(a,x^2-x+1\)

\(x^2-x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(< =>MIN=\frac{3}{4}\)dấu"=" xảy ra khi \(x=\frac{1}{2}\)

\(b,x^2+y^2-4\left(x+y\right)+16\)

\(x^2+y^2-4x-4y+16\)

\(\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+8\)

\(\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)

\(MIN=8\)dấu "=" xảy ra  khi \(x=y=2\)

\(2x^2+8x+9\)

\(\left(x^2+8x+16\right)+x^2-7\)

\(\left(x+4\right)^2+x^2-7\ge-7\)

\(< =>MIN=-7\)dấu "=" xảy ra khi \(x=-4\)