\(a,3\cdot x-15=x+35\)

\(\Rightarrow3x-x=35+15\)

\(\Rightarrow 2x=50\)

\(\Rightarrow x = 50:2\)

\(\Rightarrow x= 25\)

\(b,(8x-16)(x-5)=0\)

\(+, TH1: 8x-16=0\)

\(\Rightarrow8x=16\)

\(\Rightarrow x = 16:8\)

\(\Rightarrow x=2\)

\(+,TH2: x-5=0\)

\(\Rightarrow x =5\)

\(c,x(x+1)=2+4+6+8+10+...+2500\)  \(^{\left(1\right)}\)

Đặt \(A=2+4+6+8+10+...+2500\)

Số các số hạng của \(A\) là: \(\left(2500-2\right):2+1=1250\left(số\right)\)

Tổng \(A\) bằng: \(\left(2500+2\right)\cdot1250:2=1563750\)

Thay \(A=1563750\) vào \(^{\left(1\right)}\), ta được:

\(x\left(x+1\right)=1563750\)

\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)

\(\Rightarrow x =1250\)

#\(Toru\)

1 .x+5  và 2y+1 là Ư(42) lập bảng tính

2.vd tc chia hết 

Đề

`<=> (x-5)^15 - (x-5)^5 = 0`

`<=> (x-5)^5 . ((x-5)^10 - 1) = 0`

`<=> (x-5)^5 = 0` hoặc `(x-5)^10 - 1 = 0`

`<=> x-5 = 0` hoặc `(x-5)^10 = 1`

`<=> x = 5` hoặc `x-5 = 1` hoặc `x - 5 = -1`

`<=> x = 5` hoặc `x = 6` hoặc `x = 4` (ko t/m)

Vậy `x = 5` hoặc `x = 6`

\(\left(x-5\right)^5=\left(x-5\right)^{15}\\ \Rightarrow\left(x-5\right)^5-\left(x-5\right)^{15}=0\\ \Rightarrow\left(x-5\right)^5\left[1-\left(x-5\right)^{10}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^5=0\\1-\left(x-5\right)^{10}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{10}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(T/m\right)\\x=6\left(T/m\right)\\x=-4\left(L\right)\end{matrix}\right.\)

Lời giải:
a.

$(x-15).27=0$

$x-15=0:27=0$

$x=15+0=15$

b.

$23(42-x)=0$

$42-x=0$

$x=42$

c.

$(9x+2).3=60$

$9x+2=60:3=20$

$9x=18$

$x=2$
d.

$71+(26-3x):5=75$

$(26-3x):5=75-71=4$

$26-3x=4.5=20$

$3x=26-20=6$

$x=6:2=3$

\(\Rightarrow\left(x+5\right)\left(y-3\right)=1\cdot15=3\cdot5\)

Ta có 

Vậy \(\left(x;y\right)=\left\{\left(10;4\right);\left(0;6\right)\right\}\)

\(\sqrt{4\left(x+1\right)}=\sqrt{8}\)

⇒4(x+1)=8

⇒x+1=2

⇒x=1

a. \(\sqrt{4\left(x+1\right)}=\sqrt{8}\)                    ĐKXĐ: \(x\ge-1\)

<=> \(\left(\sqrt{4\left(x+1\right)}\right)^2=\left(\sqrt{8}\right)^2\)

<=> 4(x + 1) = 8

<=> 4x + 4 = 8

<=> 4x = -4

<=> x = -1 (TM)

Vậy nghiệm của PT là S = \(\left\{-1\right\}\)

d) Ta có: \(n^2+5n+9⋮n+3\)

\(\Leftrightarrow n^2+3n+2n+6+3⋮n+3\)

\(\Leftrightarrow n\left(n+3\right)+2\left(n+3\right)+3⋮n+3\)

mà \(n\left(n+3\right)+2\left(n+3\right)⋮n+3\)

nên \(3⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(3\right)\)

\(\Leftrightarrow n+3\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{-2;-4;0;-6\right\}\)

Vậy: \(n\in\left\{-2;-4;0;-6\right\}\)

d) Ta có: n2+5n+9⋮n+3n2+5n+9⋮n+3

⇔n2+3n+2n+6+3⋮n+3⇔n2+3n+2n+6+3⋮n+3

⇔n(n+3)+2(n+3)+3⋮n+3⇔n(n+3)+2(n+3)+3⋮n+3

mà n(n+3)+2(n+3)⋮n+3n(n+3)+2(n+3)⋮n+3

nên 3⋮n+33⋮n+3

⇔n+3∈Ư(3)⇔n+3∈Ư(3)

`b)` - Ta thấy : `|x+1|+|x-2|+|x+7|>=0`

`-> 5x-10>=0`

`-> 5x>=10`

`-> x>=2`

`-> |x+1|=x+1;|x-2|=x-2;|x+7|=x+7`

- Vậy ta có :

`(x+1)+(x-2)+(x+7)=5x-10`

`<=> x+1+x-2+x+7=5x-10`

`<=> 3x+6=5x-10`

`<=> 3x-5x=-10-6`

`<=> -2x=-16`

`<=> x=8`

a)

\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)

Vậy \(x = 2\)

b)

\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)

Vậy \(x = 3\)

Lời giải:
a.

$0< x< \frac{1}{4}+\frac{4}{5}$

$\Rightarrow 0< x< \frac{21}{20}$ hay $0< x< 1,05$

$\Rightarrow x=1$

b.

$\frac{4}{7}+\frac{3}{7}< x< \frac{5}{3}+\frac{2}{3}$
$\Rightarrow 1< x< \frac{7}{3}$
$\Rightarrow x=2$