ta có 5^x+5^x*25=650 suy ra 5^x *26=650 ,5^x=25 suy ra x=2

Mk giúp phần đầu nhé!

Có 5^x+5^x+2=650

=> 5^x.(1+5²)=650

     5^x. 26=650

     5^x=650:26

     5^x=25

=> x=2

A=0 rồi mà bạn

bai EZ​ qua

????

:))))

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a: \(\left(1-\frac12\right)\times\left(1-\frac13\right)\times\ldots\times\left(1-\frac{1}{99}\right)\)

\(=\frac12\times\frac23\times\ldots\times\frac{98}{99}\)

\(=\frac{1}{99}\)

b: \(\left(x+\frac12\right)+\left(x+\frac16\right)+\left(x+\frac{1}{12}\right)+\left(x+\frac{1}{20}\right)+\cdots+\left(x+\frac{1}{90}\right)=\frac{99}{10}\)

=>\(\left(x+\frac{1}{1\times2}\right)+\left(x+\frac{1}{2\times3}\right)+\cdots+\left(x+\frac{1}{9\times10}\right)=\frac{99}{10}\)

=>\(9\times x+1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x+1-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x=\frac{99}{10}+\frac{1}{10}-1=\frac{100}{10}-1=10-1=9\)

=>x=1

1) 5 + (-4) = 1

2) (-8) + 2 = -6

3) 8 + (-2) = 6

4) 11 + (-3) = 8

5) (-11) + 2 = -9

6) (-7) + 3 = -4

7) (-5) + 5 = 0

8) 11 + (-12) = -1

9) (-18) + 20 = 2

10) (15) + (-12) = 3

11) (-17) + 17 = 0

12) 16 + (-2) = 14

13) (30) + (-14) = 16

14) (-19) + 20 = 1

15) (-18) + 15 = -3

16) (10) + (-6) = 4

17) (-28) + 14 = -14

18) 15 + (-30) = -15

19) (15) + (-4) = 11

20) (-21) + 11 = -10

21) 8 + (-22) = -14

22) (-15) + 4 = -11

23) (-3) + 2 = -1

24) 17 + (-14) = 3

25) 17 + (-14) = 3

thế này đọc đc mới đỉnh

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

$(x-1)(x^2+x+1)-x^3-6x=11$

Dùng $(x-1)(x^2+x+1)=x^3-1$:

$x^3-1-x^3-6x=11$

$-6x-1=11$

$-6x=12$

$x=-2$

Vậy $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

Điều kiện: $x\ne-2,-1$.

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

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

$-4x=5$

$x=-\dfrac54$

Vậy $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

Điều kiện: $x\ne-2,-1$.

$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$

$x+1+2x+4=0$

$3x+5=0$

$x=-\dfrac53$

Vậy $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

Điều kiện: $x\ne0,3$.

$\dfrac{9-x^2}{x(x-3)}=1$

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

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

$2x^2-3x-9=0$

$(2x+3)(x-3)=0$

$x=-\dfrac32$ hoặc $x=3$

Nhưng $x=3$ không thỏa điều kiện.

Vậy $x=-\dfrac32$.