Bài 1: Tìm x, biết :
e, ( x - 5) . 30% = 200% x + 5
g, x /1.2 + x /2.3+... + x / 2003.2004 =1
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2 ) 1.2 + 2.3 + 3.4 + ... + x.( x + 1 ) = 5850
Đặt A = 1.2 + 2.3 + 3.4 + ... + x.( x + 1 )
=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.(5 - 2 ) + ... + x.( x + 1 ). [ ( x + 2 ) - ( x - 1 ) ]
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + x . ( x + 1 ) . ( x + 2 ) - ( x - 1 ). x . ( x + 1 )
= x . ( x + 1 ) . ( x + 2 )
=> A = [ x . ( x + 1 ) . ( x + 2 ) ] : 3
Mà tổng A = 5850
=> [ x . ( x + 1 ) . ( x + 2 ) ] : 3 = 5850
=> x . ( x + 1 ) . ( x + 2 ) = 17550
x . ( x + 1 ) . ( x + 2 ) = 25 . 26 . 27
=> x = 25
Vậy x = 25
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
\(\left(d\right)30-x=200-175\)
\(30-x=25\)
\(x=5\)
\(\left(e\right)5\left(x+12\right)=70\)
\(x+12=14\)
\(x=2\)
tương tự 2 câu trên bạn tự làm câu cuối nhé, mình không ghi lại đề đâu
#quankun^^
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)
\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)
\(\text{Vậy }x=64\)
(X-5)×\(\dfrac{30}{100}\)=\(\dfrac{2000}{100}\)x+5
(X-5)×\(\dfrac{3}{10}\)=2x+5
3/10x-15/10=2x+5
3/10x-(-2x)=15/10+5
3/10x+2x=13/2
23/10x=13/2
X=13/2÷23/10
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