3\(^{x+1}\) - 3\(^x\) = 1428

3\(^x\).( 3 - 1) = 1428

3\(^x\).2        = 1428

3\(^x\)           = 1428: 2

3\(^x\)           = 714 

3\(^{x+1}\) là số lẻ \(\forall\) \(x\) ⇒ 3\(^x\) \(\ne\) 714 ∀ \(x\) ⇒ \(x\) \(\in\) \(\varnothing\)

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

Lời giải:

a. $x=|x+1|+|x+2|+|x+3|\geq 0$

$\Rightarrow x+1>0; x+2>0; x+3>0$

$\Rightarrow |x+1|=x+1; |x+2|=x+2; |x+3|=x+3$. Do đó:

$(x+1)+(x+2)+(x+3)=x$

$3x+6=x$

$2x+6=0$

$x=-3< 0$ (vô lý)

Vậy pt vô nghiệm.

b.

$|2x+1|\geq 0$

$|x-y+1|\geq 0$

Do đó để tổng của chúng bằng $0$ thì:

$2x+1=x-y+1=0$

$\Leftrightarrow x=\frac{-1}{2}; y=\frac{1}{2}$

c.

$|x-3|=x-3$

$\Leftrightarrow x\geq 3$

c: Ta có: \(\left|x-3\right|+3=x\)

\(\Leftrightarrow\left|x-3\right|=x-3\)

\(\Leftrightarrow x-3\ge0\)

hay \(x\ge3\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{1}{4}\)

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{3}-\frac{1}{4}\)

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)

\(\frac{2}{3}-x=\frac{1}{12}-\frac{5}{4}\)

\(\frac{2}{3}-x=-\frac{7}{6}\)

\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)

\(x=\frac{2}{3}+\frac{7}{6}\)

\(x=\frac{11}{6}\)

a) cho f(x) = 0

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

\(=>\left[{}\begin{matrix}x+2=0\\-x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

b)  2(x-3)-3(x+1)=5 

\(\Leftrightarrow2x-6-3x-3=5\)

\(\Leftrightarrow-x-9=5\)

\(\Leftrightarrow-x=14\Leftrightarrow x=14\)

`a)` Cho `f(x)=0`

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

`@TH1:x+2=0=>x=-2`

`@TH2:-x+1=0=>-x=-1=>x=1`

____________________________________________________

`b)2(x-3)-3(x+1)=5`

`=>2x-6-3x-3=5`

`=>2x-3x=5+6+3`

`=>-x=14`

`=>x=-14`