( 7x + 1 ) .68 = 610
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`5/2 -3(1/3-x)=1/4-7x`
`=> 5/2 - 1 + 3x=1/4 -7x`
`=>3x+7x= 1/4 - 5/2 +1`
`=> 10x= 1/4 - 10/4 +4/4`
`=>10x= -5/4`
`=>x=-5/4 :10`
`=>x=-5/4 xx1/10`
`=>x= -5/40=-1/8`
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{197}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{40}{200}-\frac{1}{200}\)
\(=\frac{39}{200}\)
#H
Trả lời:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\)
\(=\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+\frac{1.3}{11.14.3}+...+\frac{1.3}{197.200.3}\)
\(=\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{200}\right)=\frac{1}{3}.\frac{39}{200}=\frac{13}{200}\)
| 7x - 10 | + 7x = 10
| 7x - 10 | = 10 - 7x
=> 7x - 10 thuộc { 10 - 7x; -10 + 7x }
+) 7x - 10 = 10 - 7x
7x + 7x = 10 + 10
14x = 20
x = 10/7
+) 7x - 10 = -10 + 7x
7x - 7x = -10 + 10
0x = 0
=> x = 0
Học tốt O.o
TH1 : x < 10/7
=> / 7x - 10 / +7x = 10 - 7x + 7x = 10 ( luôn đúng )
=> với mọi x < 10/7, ta luôn có / 7x - 10 / + 7x = 10
TH2 : x = 10/7
=> /7x - 10/ + 7x = 0 + 7x = 10
=> x = 10/ 7 ( thỏa mãn )
TH3 : x > 10/7
=> / 7x - 10 / + 7x = 7x - 10 + 7x = 10
=> 14 x = 10
=> x = 10/ 7 ( loại )
Vậy với mọi x < 10/7, biểu thức trên luôn đúng
\(50\%+\dfrac{1}{2}+\dfrac{1}{7}\times\dfrac{2}{5}+\dfrac{1}{7}\times\dfrac{3}{5}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{2}{35}+\dfrac{3}{35}\)
\(=1+\dfrac{5}{35}\)
\(=1+\dfrac{1}{7}\)
\(=\dfrac{8}{7}\)
\(50\%+\dfrac{1}{2}+\dfrac{1}{7}x\dfrac{2}{5}+\dfrac{1}{7}x\dfrac{3}{5}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{7}x\left(\dfrac{2}{5}+\dfrac{3}{5}\right)\)
\(=1+\dfrac{1}{7}x1\)
\(=\dfrac{7}{7}+\dfrac{1}{7}=\dfrac{8}{7}\)
50% + 1/2 + 1/7 × 2/5 + 1/7 × 3/5
= 1/2 + 1/2 + 1/7 × (2/5 + 3/5)
= 1 + 1/7 × 1
= 1 + 1/7
= 7/7 + 1/7
= 8/7
Ừ có báo cáo nói chuyện nha!
@congtybaocao
\(3\cdot\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{1}{10}\right)=-\dfrac{7}{4}\\ \Leftrightarrow-8x=-\dfrac{15}{4}\\ \Leftrightarrow x=\dfrac{15}{32}\)
\(4\left(\dfrac{3}{4}+x\right)-\dfrac{1}{5}=7x-\dfrac{1}{7}\\ \Leftrightarrow3x=\dfrac{103}{35}\\ \Leftrightarrow x=\dfrac{103}{105}\)
\(3\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{1}{10}\right)=-\dfrac{7}{4}\)
\(\Rightarrow\dfrac{3}{2}-3x-5x+\dfrac{1}{2}=-\dfrac{7}{4}\)
\(\Rightarrow2-8x=-\dfrac{7}{4}\)
\(\Rightarrow8x=2+\dfrac{7}{4}\)
\(\Rightarrow8x=\dfrac{15}{4}\)
\(\Rightarrow x=\dfrac{15}{32}\)
___________
\(4\left(\dfrac{3}{4}+x\right)-\dfrac{1}{5}=7x-\dfrac{1}{7}\)
\(\Rightarrow3+4x-\dfrac{1}{5}=7x-\dfrac{1}{7}\)
\(\Rightarrow4x+\dfrac{14}{5}=7x-\dfrac{1}{7}\)
\(\Rightarrow7x-4x=\dfrac{14}{5}+\dfrac{1}{7}\)
\(\Rightarrow3x=\dfrac{103}{35}\)
\(\Rightarrow x=\dfrac{103}{105}\)
(7x+1).68=610
=> 7x+1=62=36
=> 7x=35=> x=5
Tìm x:( 7x + 1 ) .68 = 610
Giải:Ta có:\(\left(7x+1\right).6^8=6^{10}\)
\(\Rightarrow7x+1=\frac{6^{10}}{6^8}\Rightarrow7x+1=6^2\)
\(\Rightarrow7x+1=36\Rightarrow7x=35\Rightarrow x=5\)
Vậy x=5 thỏa mãn