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Ta có:
A =2100-299+298-297+.....+22-21
=>2A=2101-2100+299-298+.....+23-22
=>2A+A=(2101-2100+299-298+.....+23-22) + (2100-299+298-297+....+22-21)
=>3A=2101-2
=>A=\(\frac{2^{101}-2}{3}\)
Vậy A=\(\frac{2^{101}-2}{3}\).
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(\Rightarrow3A=2^{101}-2\)
\(\Rightarrow A=\frac{2^{101}-2}{3}\)
!)
=> x(x - 1)=0
=> \(\left[\begin{array}{nghiempt}x=1\\x-1=0\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
Vậy đa thức có nghiệm là x=0 ; x=1
1) \(x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
b) \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
c)\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\end{array}\right.\)
d)\(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\3x-4=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{4}{3}\end{array}\right.\)
. Chú ý dấu chấm là nhân nha.
a: \(=2016+\dfrac{\dfrac{1}{5}+\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{10}-\dfrac{15}{22}}=2016+\dfrac{453}{440}:\dfrac{-9}{110}\)
\(=2016-\dfrac{151}{12}=\dfrac{24343}{12}\)
b: \(=\dfrac{1,3-13.2}{2.6}-\dfrac{5}{6}:2\)
\(=\dfrac{-119}{26}-\dfrac{5}{12}=\dfrac{-779}{156}\)
c: \(=15\left(-1-\dfrac{5}{7}-\dfrac{2}{7}\right)+\left(-105\right)\cdot\dfrac{1}{105}\)
\(=-30-1=-31\)
b)(x+a)(x+b)(x+c)=x2+(a+b+c)x2+(ab+bc+ac)x+abc
(x2+bx+ax+ab)(x+c)=x3+ax2+bx2+cx2+abx+bcx+acx+abc
x3+ax2+bx2+cx2+abx+bcx+acx+abc=x3+ax2+bx2+cx2+abx+bcx+acx+abc(1)
Vì hai biểu thức trên (1) giông nhau
Do đó (x+a)(x+b)(x+c)=x2+(a+b+c)x2+(ab+bc+ac)x+abc
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)
a/ theo bài ra, ta có:
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z\)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+z+x+1+x+y-2}=\frac{x+y+z}{2\left(x+y+z\right)}=x+y+z\)
- nếu x+y+z = 0 => x = y= z = 0
- nếu x+y+z khác 0 => x+y+z = \(\frac{1}{2}\)
=> y + z = \(\frac{1}{2}\) - x
=> z + x = \(\frac{1}{2}\) - y
=> x + y = \(\frac{1}{2}\) - z
=> \(\frac{x}{\frac{1}{2}-x+1}=\frac{y}{\frac{1}{2}-y+1}=\frac{z}{\frac{1}{2}-z-2}=\frac{1}{2}\)
=> 2x = \(\frac{1}{2}\) - x + 1 => x = \(\frac{1}{2}\)
=> 2y = \(\frac{1}{2}-y+1\) => y = \(\frac{1}{2}\)
=> 2z = \(\frac{1}{2}-z-2\) => z = \(\frac{-1}{2}\)
vậy x = 0 hoặc 1/2
y = 0 hoặc 1/2
z = 0 hoặc -1/2
mk lm câu b bái 1 nha
Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\=\frac{2x+3y-z-2-6+3}{9}=\frac{2x+3y-z-5}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)
Suy ra
x - 1 = 5 . 2 = 10
x = 10 + 1
→ x = 11
y - 2 = 3 . 5 = 15
y = 15 + 2
→ y = 17
z - 3 = 4 . 5 = 20
z = 20 + 3
→ z = 23
a) A = 1 + 2 + 22 + 23 + ... + 22012
2A = 2 + 22 + 23 + 24 + ... + 22013
2A - A = (2 + 22 + 23 + 24 + ... + 22013) - (1 + 2 + 22 + 23 + ... + 22012)
A = 22013 - 1
b) A = 22013 - 1
A = 22012.2 - 1
A = (24)503.2 - 1
A = (...6)503.2 - 1
A = (...6).2 - 1
A = (...2) - 1
A = (...1)
c) A = 1 + 2 + 22 + 23 + ... + 22012 (có 2013 số; 2013 chia hết cho 3)
A = (1 + 2 + 22) + (23 + 24 + 25) + ... + (22010 + 22011 + 22012)
A = 7 + 23.(1 + 2 + 22) + ... + 22010.(1 + 2 + 22)
A = 7 + 23.7 + ... + 22010.7
\(A=7.\left(1+2^3+...+7^{2010}\right)⋮7\left(đpcm\right)\)
cảm ơn bn đã giúp mk bài này nha. Mk cn một bài nữa giúp mk nốt nha cảm ơn bn nhìu