\(x+\left(x+2\right)+\left(x+4\right)+.....+\left(X+98\right)+\left(x+100\right)=3060\)
Bt thực hiện phép tính:
a/\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\)
b/\(\dfrac{1}{x\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)\left(x+6\right)}+...+\dfrac{1}{\left(x+96\right)\left(x+98\right)\left(x+100\right)}\)
GIÚP MÌNH VỚI!!!!!! THỨ HAI THẦY KIỂM TRA RỒI!
\(A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(A=\dfrac{1}{x}-\dfrac{1}{x+2018}=\dfrac{2018}{x\left(x+2018\right)}\)
\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}-\dfrac{1}{\left(x+4\right)\left(x+6\right)}+...+\dfrac{1}{\left(x+96\right)\left(x+98\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)\)
\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)=\dfrac{1}{4}\left(\dfrac{x^2+198x+9800-x^2-2x}{x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\right)\)
\(B=\dfrac{196x+9800}{4x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=-2\dfrac{1}{2}\)
\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{-2}{3}\right)^{98}\)
\(\dfrac{3}{2}:\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{4}\)
a: \(\dfrac{3}{4}+\dfrac{1}{4}:x=-2\dfrac{1}{2}\)
=>\(\dfrac{1}{4}:x=-\dfrac{5}{2}-\dfrac{3}{4}=-\dfrac{10}{4}-\dfrac{3}{4}=-\dfrac{13}{4}\)
=>\(x=\dfrac{-1}{4}:\dfrac{13}{4}=\dfrac{-1}{4}\cdot\dfrac{4}{13}=\dfrac{-1}{13}\)
b: \(\left(\dfrac{2}{3}\right)^{100}:x=\left(-\dfrac{2}{3}\right)^{98}\)
=>\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{2}{3}\right)^{98}\)
=>\(x=\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{2}{3}\right)^{98}=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
c: \(\dfrac{3}{2}:\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{4}\)
=>\(\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{2}:\dfrac{3}{4}=\dfrac{3}{2}\cdot\dfrac{4}{3}=2\)
=>\(\left[{}\begin{matrix}4x-\dfrac{1}{5}=2\\4x-\dfrac{1}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{5}\\4x=-\dfrac{9}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{11}{20}\\x=-\dfrac{9}{20}\end{matrix}\right.\)
\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy...
c)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=90\)
d)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+.....+\left(x+99\right)+\left(x+100\right)+=20150\)
c) (x+1) + (x+2) + ... + (x+5) = 90
=> 5x + ( 1 + 2 + ... + 5 ) = 90
5x + 15 = 90
5x = 90 - 15
5x = 75
x = 75 : 5
x = 15
d) (x+1) + (x+2) + .... + (x+100) = 20150
=> 100x + ( 1+2+...+100 ) = 20150
100x + 5050 = 20150
100x = 20150 - 5050
100x = 15100
x = 15100 : 100
x = 151
Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90
<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90
<=> 5x + 15 = 90
=> 5x = 75
=> x = 15
c) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=90\)
\(\Leftrightarrow x+1+x+2+x+3+x+4+x+5=90\)
\(\Leftrightarrow5x+\left(1+2+3+4+5\right)=90\)
\(\Leftrightarrow5x+15=90\)
\(\Leftrightarrow5x=75\)
\(\Leftrightarrow x=15\)
d) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+......+\left(x+99\right)+\left(x+100\right)=20150\)
\(\Leftrightarrow x+1+x+2+x+3+......+x+99+x+100=20150\)
\(\Leftrightarrow100x+\left(1+2+3+.....+99+100\right)=20150\)
\(\Leftrightarrow100x+5050=20150\)
\(\Leftrightarrow100x=15100\)
\(\Leftrightarrow x=151\)
tìm x \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+...+\left(x+99\right)+\left(x+100\right)=5750\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(\left(x\cdot100\right)+\left(1+2+...+100\right)=5750\)
\(\left(x\cdot100\right)+\left(100+1\right)\cdot\frac{100}{2}=5750\)
\(\left(x\cdot100\right)+101\cdot50=5750\)
\(\left(x\cdot100\right)+5050=5750\)
\(x\cdot100=5750-5050\)
\(x\cdot100=700\)
\(x=700\div100\)
\(x=7\)
Ta có: ( x+1)+(x+2)+(x+3)+.....+(x+99)+(x+100)=5750
<=>(x+x+x+....+x+x)+(1+2+3+..+99+100)=5750
<=> 100x+5050=5750
=>100x=5750-5050
=>100x=700
=>x=700:100
=>x=7
Vậy x=7
hoặc mở câu hỏi tương tự tham khảo.
câu 1: giải hệ phương trình
\(\left(x+y\right)^2+\left(y+z\right)^4+....+\left(x+z\right)^{100}=-\left(y+z+x\right)\)
\(\left(xy\right)^2+2\left(yz\right)^4+....+100\left(zx\right)^{100}=-[\left(x+y+z\right)+2\left(yz+zx+xy\right)+......+99\left(x+y+z\right)]\)\(\left(\frac{1}{x}+\frac{1}{y}\right)^2+\left(\frac{1}{y^2}+\frac{1}{z^2}\right)^2+...+\left(\frac{1}{x^{99}}+\frac{1}{z^{99}}\right)^2=-\frac{1}{\left(xy\right)^2+2\left(yz\right)^2+.....+99\left(zx\right)^2}\)
tìm x,y,z
Đúng là chơi lừa bịp thực sự bài này rất dễ đây là cách giải:
ta có: \(\left(x+y\right)^2+\left(y+z\right)^4+.....+\left(x+z\right)^{100}\ge0\)còn \(-\left(y+z+x\right)\le0\) nên phương trình 1 vô lý
tương tự chứng minh phương trinh 2 và 3 vô lý
vậy \(\hept{\begin{cases}x=\varnothing\\y=\varnothing\\z=\varnothing\end{cases}}\)
thực sự bài này mới nhìn vào thì đánh lừa người làm vì các phương trình rất phức tạp nhưng nếu nhìn kĩ lại thì nó rất dễ vì các trường hợp đều vô nghiệm
\(\left(x+y\right)^2+\left(y+z\right)^4+...+\left(x+z\right)^{100}=-\left(y+z+x\right)\)
Đặt : \(A=\left(x+y\right)^2+\left(y+z\right)^4+...+\left(x+z\right)^{100}\)
Ta dễ dàng nhận thấy tất cả số mũ đều chẵn
\(=>A\ge0\)(1)
Đặt : \(B=-\left(y+z+x\right)\)
\(=>B\le0\)(2)
Từ 1 và 2 \(=>A\ge0\le B\)
Dấu "=" xảy ra khi và chỉ khi \(A=B=0\)
Do \(B=0< =>y+z+x=0\)(3)
\(A=0< =>\hept{\begin{cases}x+y=0\\y+z=0\\x+z=0\end{cases}}\)(4)
Từ 3 và 4 \(=>x=y=z=0\)
Vậy nghiệm của pt trên là : {x;y;z}={0;0;0}
Đặt :\(\left(xy\right)^2+2\left(yz\right)^4+...+100\left(zx\right)^{100}=A\)
Ta thấy các số mũ đều chẵn
Nên \(A\ge0\left(1\right)\)
Đặt : \(-\left[\left(x+y+z\right)+2\left(yz+zx+xy\right)+...+99\left(x+y+z\right)\right]=B\)
Vì có dấu âm ở trước VT
Nên \(B\le0\left(2\right)\)
Từ 1 và 2 <=> \(A=B=0\)
\(< =>x=y=z=0\)
Thu gọn đa thức \(P\left(x\right)=x^2+\left(x+2\right)^2+\left(x+3\right)^2+...+\left(x+98\right)^2-\left[\left(x+1\right)^2+\left(x+3\right)^2+...+\left(x+99\right)^2\right]\)
đc đa thức P(x) = ax + b vậy a - b là
Ta có:
P(x)=x2+(x+2)2+(x+3)2+...+(x+98)2−[(x+1)2+(x+3)2+...+(x+99)2]
=[x2-(x+1)2]+[(x+2)2-(x+3)2]+[(x+3)2-(x+4)2]+...+[(x+98)2-(x+99)2]
=(x-x-1)(x+x+1)+(x+2-x-3)(x+2+x+3)+...+(x+98-x-99)(x+98+x+99)
=-(2x+1)-(2x+5)-....-(2x+197)
=(-2x-2x-...-2x)+(-1-5-...-197)
Vì đa thức trên có \(\dfrac{197-1}{4}+1=50\text{ số hạng => -2x có 50 hạng tử}\)
Nên ta có:
=(-2x*50)+\(\left(\dfrac{\left(-197-1\right)\cdot50}{2}\right)\)
=-100x-4950
Mà P(x)=ax+b =>{a=-100; b=-4950}
Vậy a-b= -100-(-4950)= 4850 (Hihi! Mình tự làm nên ko biết đúng hay ko?)
Rút gọn các biểu thức sau :
a) \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3-x^3-3x\left(x+1\right)\left(x-1\right)\)
c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)
d) \(100^2-99^2+98^2+97^2+......+2^2-1^2\)
e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
f) \(\left(a+b+c\right)^{^{ }2}+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
a,b,c,f tìm cách áp dụng HĐT vào nhé! động não tí xem :)
d) Sửa đề :\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=199+195+...+3\)
Khi đó tổng sẽ là:
\(\dfrac{\left(199+3\right)\left[\dfrac{\left(199-3\right)}{4}+1\right]}{2}=5050.\)
e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1\)
\(=2^{128}.\)
\(I\)Tính nhanh
\(a.127^2+146.127+73^2\)
\(b.9^8.2^8\left(18^4-1\right)\left(18^4+1\right)\)
\(c.100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(d.\frac{780^2-220^2}{125^2+150.125+75^2}\)
\(II.\)Rút gọn các biểu thức
\(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2