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A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)
[x+1]:25=3
x+1=25*3
x+1=75
x=75-1=74
12:[x-3]=6x-3=12:6
x-3=2
x=3+2
x=5
4[x-1]:5=8
4[x-1]=5.8
4[x-1]=40
x-1=40:4
x-1=10
x=10+1=11
9[x+3]-25=11
9[x+3]=25+11=36
x+3=36:9=4
x=4-3=1
75:[x-2]=1
x-2=75:1=75
x=75+2=77
[2x-5]-7=6
2x-5=7+6=13
2x=13+5=18
x=18:2=9
k mk nha mk cầu xin bn đó công mk làm mà
Dễ thế mà không làm được thì bạn nên xem lại nhé,một hai câu thì còn được chứ cả 10 câu thế kia rõ là ỷ lại rồi bạn ạ.Thân!
a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)
\(\Leftrightarrow3x+15-3=2x+2+7\)
\(\Leftrightarrow3x+12=2x+9\)
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
c) Ta có: \(75-\left(2x+1\right)^3=11\)
\(\Leftrightarrow\left(2x+1\right)^3=64\)
\(\Leftrightarrow2x+1=4\)
hay \(x=\dfrac{3}{2}\)
a) Ta có: 3(x+5)−3=2(x+1)+73(x+5)−3=2(x+1)+7
⇔3x+15−3=2x+2+7⇔3x+15−3=2x+2+7
⇔3x+12=2x+9⇔3x+12=2x+9
hay x=-3
b) Ta có: 15−(x+2)2=−115−(x+2)2=−1
⇔(x+2)2=16⇔(x+2)2=16
⇔[x+2=4x+2=−4⇔[x=2x=−6
`3(x+5)-3=2(x+1)+7`
`3x+15-3=2x+2+7`
`x=-3`
.
`15-(x+2)^2=-1`
`(x+2)^2=16`
`(x+2)^2=4^2=(-4)^2`
`[(x+2=4),(x+2=-4):}`
`[(x=2),(x=-6):}`
.
`75-(2x+1)^3=11`
`(2x+1)^3=64`
`(2x+1)^2=4^3`
`2x+1=4`
`x=3/2`
3(x+5)−3=2(x+1)+73(x+5)-3=2(x+1)+7
3x+15−3=2x+2+73x+15-3=2x+2+7
x=−3x=-3
.
15−(x+2)2=−115-(x+2)2=-1
(x+2)2=16(x+2)2=16
(x+2)2=42=(−4)2(x+2)2=42=(-4)2
[x+2=4x+2=−4[x+2=4x+2=-4
[...


\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{11}{75}\left(x\ne0;x\ne-2\right)\)
\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{22}{75}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{22}{75}\)
\(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{22}{75}\)
\(\dfrac{1}{x+2}=\dfrac{1}{3}-\dfrac{22}{75}\)
\(\dfrac{1}{x+2}=\dfrac{1}{25}\)
suy ra
\(x+2=25\\ x=23\left(tm\right)\)
=>2/3*5+2/5*7+...+2/x(x+2)=22/75
=>1/3-1/5+1/5-1/7+...+1/x-1/x+2=22/75
=>1/3-1/x+2=22/75
=>1/x+2=25/75-22/75=3/75=1/25
=>x=23
Mỗi phân số có dạng tổng quát là 1n(n+2)the fraction with numerator 1 and denominator n open paren n plus 2 close paren end-fraction1𝑛(𝑛+2). Ta có công thức tách phân số:
1n(n+2)=12(1n−1n+2)the fraction with numerator 1 and denominator n open paren n plus 2 close paren end-fraction equals one-half open paren 1 over n end-fraction minus the fraction with numerator 1 and denominator n plus 2 end-fraction close paren1𝑛(𝑛+2)=121𝑛−1𝑛+2 Bước 2: Áp dụng vào biểu thức
Vế trái của phương trình trở thành:
12(13−15)+12(15−17)+12(17−19)+…+12(1x−1x+2)one-half open paren one-third minus one-fifth close paren plus one-half open paren one-fifth minus one-seventh close paren plus one-half open paren one-seventh minus one-nineth close paren plus … plus one-half open paren 1 over x end-fraction minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren1213−15+1215−17+1217−19+…+121𝑥−1𝑥+2 Bước 3: Rút gọn vế trái
Đặt 12one-half12làm nhân tử chung:
12(13−15+15−17+17−19+…+1x−1x+2)one-half open paren one-third minus one-fifth plus one-fifth minus one-seventh plus one-seventh minus one-nineth plus … plus 1 over x end-fraction minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren1213−15+15−17+17−19+…+1𝑥−1𝑥+2Các số hạng ở giữa triệt tiêu nhau, chỉ còn lại số hạng đầu và số hạng cuối:
12(13−1x+2)=1175one-half open paren one-third minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren equals 11 over 75 end-fraction1213−1𝑥+2=1175 Bước 4: Giải phương trình tìm xx𝑥
Nhân cả hai vế với 2:
13−1x+2=2275one-third minus the fraction with numerator 1 and denominator x plus 2 end-fraction equals 22 over 75 end-fraction13−1𝑥+2=2275Chuyển vế:
1x+2=13−2275the fraction with numerator 1 and denominator x plus 2 end-fraction equals one-third minus 22 over 75 end-fraction1𝑥+2=13−2275 1x+2=2575−2275the fraction with numerator 1 and denominator x plus 2 end-fraction equals 25 over 75 end-fraction minus 22 over 75 end-fraction1𝑥+2=2575−2275 1x+2=375the fraction with numerator 1 and denominator x plus 2 end-fraction equals 3 over 75 end-fraction1𝑥+2=375 1x+2=125the fraction with numerator 1 and denominator x plus 2 end-fraction equals 1 over 25 end-fraction1𝑥+2=125 Suy ra:
x+2=25x plus 2 equals 25𝑥+2=25 x=23x equals 23𝑥=23 Vậy 𝑥 =23.