(x+1)+(x+1)+(x+1)+(x+1)+(x+1)=55
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\(\left(1-2x\right).5^4=3.5^5\)
\(1-2x=3.5^5:5^4\)
\(1-2x=3.5=15\)
\(2x=\left(-14\right)\)
\(x=\left(-7\right)\)
=[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: 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)
\(1,x-\left\{x-\left[x-\left(x-1\right)\right]\right\}=1\)
\(\Leftrightarrow x-\left\{x-x+1\right\}=1\)
\(\Leftrightarrow x-\left\{0+1\right\}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)
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 \(x=2\)
\(\left(x+1\right)+\left(x+1\right)+\left(x+1\right)+\left(x+1\right)+\left(x+1\right)=55\)
\(x+1+x+1+x+1+x+1+x+1=55\)
\(5x+5=55\)
\(5x=55-5\)
\(5x=50\)
\(x=50:5\)
\(x=10\)
TL:
X=10
Vì (10+1)+(10+1)+(10+1)+(10+1)+(10+1)=55
@hiếu
~HT~