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1 tháng 10 2021

máy tính hay tv đấy 

1 tháng 10 2021

Máy tính như hacker í

27 tháng 12 2018

24 - 16(x - 1/2) = 23

=> 16(x - 1/2) = 24 - 23

=> 16(x - 1/2) = 1

=> x - 1/2 = 1/16

=> x = 1/16 + 1/2

=> x = 9/16

27 tháng 12 2018

\(24-16(x-\frac{1}{2})=23\)

\(16(x-\frac{1}{2})=24-23\)

\(16(x-\frac{1}{2})=1\)

\(x-\frac{1}{2}=\frac{1}{16}\)

\(x=\frac{1}{16}+\frac{1}{2}\)

\(x=\frac{9}{16}\)

Vậy số thực x cần tìm là \(\frac{9}{16}\)

Chúc bạn hok tốt ~

3 tháng 7 2021

B1: a)Dấu hiệu: Điểm ktra môn Toán của 1 nhóm hs

b)Điểm(x)    |  7  |  8  |  9  |  10  |

   Tần số(n) |  5  |  7  |  5  |     3  |  N=20

-Nhận xét: +Có 3 bạn đạt điểm cao nhất là 10 điểm

                  +Có 5 bạn điểm thấp là 7 điểm

                  +Có 20 bạn tham gia làm bài

c)AD CT tính số TBC:

     \(\dfrac{x_1.n_1+x_2.n_2+...+x_4.n_4}{N}\)

=\(\dfrac{7.5+8.7+9.5+10.3}{20}\)

=8,3

-Mo=8

 

Bài 4: 

a) Xét ΔCAE vuông tại C và ΔDAE vuông tại D có 

BE chung

AC=AD(gt)

Do đó: ΔCAE=ΔDAE(Cạnh huyền-cạnh góc vuông)

Suy ra: \(\widehat{CAE}=\widehat{DAE}\)(hai góc tương ứng)

mà tia AE nằm giữa hai tia AC,AB

nên AE là tia phân giác của \(\widehat{CAB}\)

b) Ta có: ΔCAE=ΔDAE(cmt)

nên EC=ED(hai cạnh tương ứng)

Ta có: BC=BD(gt)

nên B nằm trên đường trung trực của CD(Tính chất đường trung trực của một đoạn thẳng)(1)

Ta có: EC=ED(cmt)

nên E nằm trên đường trung trực của CD(Tính chất đường trung trực của một đoạn thẳng)(2)

Từ (1) và (2) suy ra BE là đường trung trực của CD(đpcm)

10 tháng 4 2017

theo bài ra ta có:

\(\dfrac{6}{x+1}.\dfrac{x-1}{3}=\dfrac{6x-6}{3x+1}\\ =\dfrac{6x+2-8}{3x+1}\\ =\dfrac{2\left(3x+1\right)-8}{3x+1}\\ =2-\dfrac{8}{3x+1}\)

để \(\dfrac{6}{x+1}.\dfrac{x-1}{3}\) là số nguyên

=> \(\dfrac{8}{3x+1}\) nguyên

\(8⋮3x+1\\ \Rightarrow3x+1\inƯ_{\left(8\right)}=\left\{-1;1;2;-2;4;-4;8;-8\right\}\)

ta có bảng sau:

3x+1 1 -1 2 -2 4 -4 8 -8
3x 0 -2 1 -3 3 -5 7 -9
x 0 \(\dfrac{-2}{3}\) \(\dfrac{1}{3}\) -1 1 \(\dfrac{-5}{3}\) \(\dfrac{7}{3}\) -3

mà x là số nguyên

=> x ={0;-1;1;-3}

vậy x ={0;1;-1;-3}

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17 tháng 7 2016

\(\frac{x+3}{8}=\frac{2}{x-3}\)

\(\left(x+3\right)\times\left(x-3\right)=2\times8\)

\(x^2-3^2=16\)

\(x^2-9=16\)

\(x^2=16+9\)

\(x^2=25\)

\(x^2=\left(\pm5\right)^2\)

\(x=\pm5\)

Vậy x = 5 hoặc x = -5

 

 

17 tháng 7 2016

\(\frac{x+3}{8}=\frac{2}{x-3}\)

\(\Rightarrow\left(x+3\right)\left(x-3\right)=2.8=16\)

\(\Rightarrow x^2-3^2=16\)

\(\Rightarrow x^2=16+3^2=25\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)

24 tháng 12 2017

\(7^{2x}+7^{2x+2}=2450\)

\(7^{2x}.1+7^{2x}.7^2=2450\)

\(7^{2x}.\left(1+7^2\right)=2450\)

\(7^{2x}.\left(1+49\right)=2450\)

\(7^{2x}.50=2450\)

\(7^{2x}=2450:50\)

\(7^{2x}=49\)

\(7^{2x}=7^2\)

\(\Rightarrow2x=2\)

\(x=2:2\)

\(x=1\)

Vậy \(x=1\)

24 tháng 12 2017

\(7^{2x}+7^{2x+2}=2450\)

\(\Leftrightarrow7^{2x}+7^{2x}.7^2=2450\)

\(\Leftrightarrow7^{2x}.\left(1+7^2\right)=2450\)

\(\Leftrightarrow7^{2x}.50=2450\)

\(\Leftrightarrow7^{2x}=2450:50\Leftrightarrow7^{2x}=49\)

\(\Leftrightarrow7^{2x}=7^2\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

17 tháng 2 2017

a) x= 4

17 tháng 2 2017

TA CÓ \(\frac{9^X}{3^X}\)=81

=>\(3^X\)=81

=>X=4

B,

19 tháng 9 2016

Ta có :

\(xy=x:y\)

\(\Rightarrow y^2=1\)

\(\Rightarrow\left[\begin{array}{nghiempt}y=1\\y=-1\end{array}\right.\)

(+) y = 1

\(\Rightarrow x+1=x\) ( vô lý )

(+) \(y=-1\)

\(\Rightarrow x=\frac{1}{2}\) ( Nhận )

Vậy \(\left(x;y\right)=\left(\frac{1}{2};-1\right)\)

19 tháng 9 2016

thanks bạn nhìu nhìu nhayeu