tìm x : x+(x+1)+(x+2)+...+x=5353
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VT = 101x + 1+2+3+...+100 = 101x + 1/2*100*101
PT <=> 101x + 50*101 = 5353
<=> x + 50 = 53
<=> x = 3
Ta có: x+(x+1)+(x+2)+...+(x+100)=5353
<=> 101x + (1 + 2 + ....+ 100) = 5353
<=> 101x + 2550 = 5353
=> 101x = 5353 - 2550
=> 101x = 2803
=> x =2803 : 101
=> x =????????????
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5353\)
\(x+\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5353\)
\(101x+\frac{\left(1+100\right)\times100}{2}=5353\)
\(101x+5050=5353\)
\(101x=5353-5050\)
\(101x=303\)
\(\Rightarrow x=3\)
x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ......... + ( x + 100 ) = 535
101 x + ( 1 + 2 + ... + 100 ) = 535
101x + 5050 = 535
101x = -4515
\(\Rightarrow x=-\frac{4545}{101}\)
a. 16 569 330
24 425 739
b. 14 711 334
24 935 708
c. 23 222 460
28 495 512
d. 12 585 992
13 837 260
a.4674 x 3545
4563 x 5353
b.4521 x 3254 = 14711334
.4387 x 5684 = 24958869
c.4436 x 5235 = 23222460
5242 x 5436 = 28495512
d.3646 x 3452 = 12585992
3252 x 4255 = 13837260
Học tốt
a) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=12\)
\(4x=12+12+8\)
\(4x=32\)
\(x=8\)
b) \(4\left(x-5\right)-\left(x-7\right)=-19\)
\(4x-20-x+7=-19\)
\(3x=-19+20-7\)
\(3x=-6\)
\(x=-2\)
c) \(7\left(x-3\right)-5\left(3-x\right)=11x-5\)
\(7\left(x-3\right)+5\left(x-3\right)=11x-5\)
\(\left(x-3\right).12=11x-5\)
\(12x-36-11x+5=0\)
\(x-31=0\)
\(x=31\)
d. Áp dụng BĐT Caushy Schwartz ta có:
\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
a: Ta có: \(M=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
b: Để M>1 thì M-1>0
\(\Leftrightarrow\dfrac{x^2-x+1}{x-1}>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
Đề bài có vấn đề nhé bạn! Chỗ cuối x+? thì mình mới làm được