56.x=1069-1013+1-1

56.x=56+1-1

56.x=55+1

56.x=56

     x=56:56

    x=1

chuc bn hoc gioi1

x=1 nha bạn

a: \(\left(2x-1\right)^5-\left(2x-1\right)^8=0\)

=>\(\left(2x-1\right)^5\cdot\left\lbrack1-\left(2x-1\right)^3\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(2x-1\right)^5=0\\ 1-\left(2x-1\right)^3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(2x-1\right)^5=0\\ \left(2x-1\right)^3=1\end{array}\right.\)

=>\(\left[\begin{array}{l}2x-1=0\\ 2x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=1\\ 2x=2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=1\end{array}\right.\)

b: (2x+1)(2x-3)<0

TH1: \(\begin{cases}2x+1\ge0\\ 2x-3\le0\end{cases}\Rightarrow\begin{cases}x\ge-\frac12\\ x\le\frac32\end{cases}\)

=>\(-\frac12\le x\le\frac32\)

TH2: \(\begin{cases}2x+1\le0\\ 2x-3\ge0\end{cases}\Rightarrow\begin{cases}2x\le-1\\ 2x\ge3\end{cases}\Rightarrow\begin{cases}x\le-\frac12\\ x\ge\frac32\end{cases}\)

=>x∈∅

c: (x-1)(2x+3)>0

TH1: \(\begin{cases}x-1>0\\ 2x+3>0\end{cases}\Rightarrow\begin{cases}x>1\\ x>-\frac32\end{cases}\)

=>x>1

TH2: \(\begin{cases}x-1<0\\ 2x+3<0\end{cases}\Rightarrow\begin{cases}x<1\\ x<-\frac32\end{cases}\)

=>\(x<-\frac32\)

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

a) \(5^{x-1}+5^{x-3}=650\)

\(\Rightarrow5^x\left(\frac{1}{5}+\frac{1}{125}\right)=650\)

\(\Rightarrow5^x=650:\frac{26}{125}\)

\(\Rightarrow5^x=3125\)

\(\Rightarrow5^x=5^5\)

\(\Rightarrow x=5\)

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).