Tìm x :
( x + 1 ) + ( x + 3 ) + ( x + 5 ) + .......... + ( x + 399 ) = 40400
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a) (2x - 10)37 = 0
2x - 10 = 0 : 37
2x - 10 = 0
2x = 0 + 10
2x = 10
x = 10 : 2
x = 5
b) 135(34 - x) = 810
34 - x = 810 : 135
34 - x = 6
x = 34 - 6
x = 28
Bài 5 :
S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399
S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)
S = 1 + 0 + 0 + ... + 0 + (- 401)
S = 1 - 401
S = - 400
Bài 5
A= 1+3-5-7+9+11-13-15+...-397-399
A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)
A= -8 -8 -...-8
A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)
A= -400
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\left(\text{Quy đồng nhé !}\right)\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{800}\)
=> x + 1 = 800
<=> x = 799
1/3+1/6+1/10+...+1/x(x+1):2=399/400
2.[1/3.2+1/6.2+1/10.2+...+1/x(x+1)]=399/400
2.[1/6+1/12+1/20+...+1/x(x+1)]=399/400
2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=399/400
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=399/800
1/2-1/x+1=399/800
1/x+1=1/800
=> x+1=800
=> x=799
Bài 2
a) \(x\times1\frac{1}{4}=3\frac{3}{4}\)
\(x\times\frac{5}{4}=\frac{15}{4}\)
\(x=\frac{15}{4}.\frac{4}{5}\)
\(x=3\)
b) \(x-\frac{3}{4}=6\times\frac{3}{8}\)
\(x-\frac{3}{4}=\frac{9}{4}\)
\(x=\frac{9}{4}+\frac{3}{4}\)
\(x=3\)
Những câu còn lại tương tự
\(K=\left(1-\dfrac{3}{2\cdot4}\right)\left(1-\dfrac{3}{3\cdot5}\right)\cdot...\cdot\left(1-\dfrac{3}{19\cdot21}\right)\)
\(=\dfrac{3^2-1-3}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2-1-3}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{20^2-4}{\left(20-1\right)\left(20+1\right)}\)
\(=\dfrac{\left(3-2\right)\left(3+2\right)}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{\left(4-2\right)\left(4+2\right)}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{18\cdot22}{\left(20-1\right)\left(20+1\right)}\)
\(=\dfrac{1\cdot5}{2\cdot4}\cdot\dfrac{2\cdot6}{3\cdot5}\cdot...\cdot\dfrac{18\cdot22}{19\cdot21}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot21\cdot22}{2\cdot3\cdot4\cdot5\cdot...\cdot19\cdot20\cdot21}=1\cdot22=22\)
(x+1)+(x+3)+(x+5)+...+(x+399)=40400
<=> x +1 + x + 3 + x + 5 + ...+ x + 399 = 40400
<=> ( x+x+x+...+x)+(1+3+5+...+399)=40400
<=> 200x + 40000=40400
<=> 200x = 400
<=> x = 2
Vậy...
\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+..+\left(x+399\right)=40400\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+3+5+..+399\right)=40400\)
\(\Leftrightarrow200.x+\left(399+1\right).200:2=40400\) ( 200 số hạng, 200 x )
\(\Leftrightarrow200x+40000=40400\)
\(\Leftrightarrow200x=400\)
\(\Leftrightarrow x=2\)
Vậy x = 2