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\(a,\left(x-\frac{1}{2}\right)^2=0\)
\(\left(x-\frac{1}{2}\right)^2=0^2\)
\(x-\frac{1}{2}=0\)
x =0+\(\frac{1}{2}\)
x =\(\frac{1}{2}\)
a) (x- 1/2)2 = 0
x-1/2 = 0
x = 1/2
c) (x+1)2 =16
(x+1)2 =42
* TH 1: x+1 =4
x = 4-1
x = 3
* TH 2 : x +1 = -4
x = -4 -1
x = -5
b) ( x-2)2 = 1
(x-2)2 = 12
Suy ra: TH 1: x-2=1
x = 3
TH 2: x-2= -1
x = 1
Đáp số : a) x=1/2
b) x=3; x=1
c) x=3; x=-5
Học tốt nha !
a,3x-10=2x+13
\(\Rightarrow\)3x-2x=10+13
\(\Rightarrow\)x=23
b,x+12=-5-x
\(\Rightarrow\)x+x=-12-5
\(\Rightarrow\)2x=-17
\(\Rightarrow\)x=-8,5
c,x+5=10-x
\(\Rightarrow\)x+x=-5+10
\(\Rightarrow\)2x=5
\(\Rightarrow\)x=2,5
e,12-x=x+1
\(\Rightarrow\)-x-x=-12+1
\(\Rightarrow\)-2x=-11
\(\Rightarrow\)x=5,5
f,14+4x=3x+20
\(\Rightarrow\)4x-3x=-14+20
\(\Rightarrow\)x=6
g,2.(x-1)+3(x-2)=x-4
\(\Rightarrow\)2x-2+3x-6=x-4
\(\Rightarrow\)2x-2+3x-6-x+4=0
\(\Rightarrow\)4x-4=0
\(\Rightarrow\)4x=4
\(\Rightarrow\)x=1
h,3(4-x)-2(x-1)=x+20
\(\Rightarrow\)12-3x-2x+2-x-20=0
\(\Rightarrow\)-6x-6=0
\(\Rightarrow\)-6x=6
\(\Rightarrow\)x=-1
i,4(2x+7)-3(3x-2)=24
\(\Rightarrow\)8x+28-9x+6=24
\(\Rightarrow\)-x+34=24
\(\Rightarrow\)-x=-10
\(\Rightarrow\)x=10
k,3(x-2)+2x=10
\(\Rightarrow\)3x-6+2x=10
\(\Rightarrow\)5x-6=10
\(\Rightarrow\)5x=16
\(\Rightarrow\)x=3,2
Phần d, tớ không biết làm!!!!
\(\left(1^5\right)^{10}=1^{5.10}=1^{50}=1\)
\(\frac{2^5}{2^3}=2^{5-3}=2^2=4\)
Vâng bây giờ đã là 29/9 r
a) x2=16
<=> x = \(\sqrt{16}\)
<=> x = 4
b) x - 12 = 4
<=> x = 4 + 1
<=> x = 5
a)\(x^2=16\)
\(\Leftrightarrow x^2=4^2\)
\(\Leftrightarrow x=4\)
vậy...
b)\(x-\left(1\right)^2=4\)
\(\Leftrightarrow x-\left(1\right)^2=\left(\pm2\right)^2\)
\(\Leftrightarrow x-1=\pm2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
vậy...
k mik nhé
\(\frac{x-2}{2}-\frac{1+x}{3}=\frac{4-3x}{4}-1\)
\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)
\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)
\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)
\(\Leftrightarrow4x-32=-18x\)
\(\Rightarrow x=\frac{16}{11}\)
làm bài & thôi :
(x2 - 2x + 3) \(⋮\)(x - 1)
= x2 - 2x + 3
=) x2 - 2x + 3 - ( x - 1 )
=) x2 - 1
=) x2 - 1 - x( x - 1 )
=) 2 \(⋮\)x - 1
tự làm
a) Ta có: (x2 - 2x + 3) \(⋮\)(x - 1)
<=> [x(x - 1) - (x - 1) + 2] \(⋮\)(x - 1)
<=> [(x - 1)2 + 2] \(⋮\)(x - 1)
Do (x - 1)2 \(⋮\)(x - 1) => 2 \(⋮\)(x - 1)
=> (x - 1) \(\in\)Ư(2) = {1; -1; 2; -2}
Lập bảng :
| x - 1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
Vậy ...
b) (3x - 1) \(⋮\)(x - 4)
<=> [3(x - 4) + 11] \(⋮\)(x - 4)
Do 3(x - 4) \(⋮\)(x - 4) => 11 \(⋮\)(x - 4)
=> (x - 4) \(\in\)Ư(11) = {1; -1; 11; -11}
Lập bảng:
| x - 4 | 1 | -1 | 11 | -11 |
| x | 5 | 3 | 15 | -7 |
vậy ...
c;d tương tự trên
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
a, x = 3 b, x=\(\varnothing\)
c, x=\(\varnothing\) d, đề bài sai
a) \(2^{x+1}=16=2^4\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
b)\(2^{3x+4}=256\)
\(\Leftrightarrow3x+4=8\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)