\(\left(1-2x\right).5^4=3.5^5\)
\(1-2x=3.5^5:5^4\)
\(1-2x=3.5=15\)
\(2x=\left(-14\right)\)
\(x=\left(-7\right)\)

\(\left(1-2x\right).5^4=3.5^5\)

\(\Rightarrow1-2x=3.\left(5^5:5^4\right)\)

\(\Rightarrow1-2x=3.5\)

\(\Rightarrow1-2x=15\)

\(\Rightarrow2x=-14\)

\(\Rightarrow x=-7\)

Vậy ...

Phần 2 là x - ( x + 2 ) > 0 nhé 

=[x(x-2)/2(x²+4)-2x²/(4+x²)(2-x)][x(x-2)(x+1)/x³]

={[x(x-2)(2-x)-4x² ]/2(2-x)(4+x²)} .[x(x-2)(x+1)/x³ ]

=[-x(x²+4)/2(2-x)(4+x²)].[x(x-2)(x+1)/x³ ]

=-x.x(x-2)(x+1)/2(2-x)x³

=(x+1)/2x

\(A=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\)

\(A=\dfrac{7}{2}\)

\(A=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\\ A=\dfrac{7}{2}\)

a=(1+1/2) x ( 1+1/3) x (1+1/4) x (1+1/5) x (1+1/6)

\(A=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}+\dfrac{1}{5}\right)\times\left(\dfrac{6}{6}+\dfrac{1}{6}\right)\)

\(A=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{7}{5}=\dfrac{7}{2}\)

a: TH1: x<3

=>3-x-2(5-x)=8

=>3-x-10+2x=8

=>x-7=8

=>x=15(loại)

TH2: 3<=x<5

=>x-3-2(5-x)=8

=>x-3-10+2x=8

=>3x=21

=>x=7(loại)

TH3: x>=5

=>x-3-2x+10=8

=>-x=1

=>x=-1(loại)

b: =>|2x-3|+3|x-4|=8x

TH1: x<3/2

=>3-2x+12-3x=8x

=>8x=-5x+15

=>13x=15

=>x=15/13(nhận)

TH2: 3/2<=x<4

=>2x-3+12-3x=8x

=>8x=-x+9

=>x=1(loại)

TH3: x>=4

=>2x-3+3x-12=8x

=>8x=5x-15

=>3x=-15

=>x=-5(loại)

ủa sao tui ko thấy gì hết vậy nè

Thay 2021= X: Thay 2022= Y
-1

\(1,x-\left\{x-\left[x-\left(x-1\right)\right]\right\}=1\)

\(\Leftrightarrow x-\left\{x-x+1\right\}=1\)

\(\Leftrightarrow x-\left\{0+1\right\}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)

Vậy x=2

\(2,|x+2|+x-1=5\)

\(\Leftrightarrow|x+2|+x=5+1\)

\(\Leftrightarrow|x+2|+x=6\)

Ta có: \(|x+2|\ge0\)

\(\Leftrightarrow x+2+x=6\)

\(\Leftrightarrow2x+2=6\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

Vậy x=2

\(x-\left\{x-\left(x-1\right)\right\}=1\)

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

\(x-1=1\)

\(x=1+1\)

\(x=2\)

Vậy \(x=2\)

\(\left|x+2\right|+x-1=5\)

\(\left|x+2\right|=5+1-x\)

\(\left|x+2\right|=6-x\)

\(\Rightarrow\orbr{\begin{cases}x+2=6-x\\x+2=-6+x\end{cases}\Leftrightarrow\orbr{\begin{cases}x+x=6-2\\x-x=-6-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\0=-8\left(v\text{ô}l\text{ý}\right)\end{cases}}}\)

Vậy \(x=2\)

vậy x=2

Ánh ak dũng đây đồng cảnh ngộ mắc bài này của cô