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LN
3
NT
2
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 ~
LA
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
4 tháng 7 2021
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)
T
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}
4 tháng 6 2021
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PA
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
\(7^{2x}+7^{2x+2}=2450\)
\(\Leftrightarrow7^{2x}+7^{2x}.7^2=2450\)
\(\Leftrightarrow7^{2x}.\left(1+7^2\right)=2450\)
\(\Leftrightarrow7^{2x}.50=2450\)
\(\Leftrightarrow7^{2x}=2450:50\Leftrightarrow7^{2x}=49\)
\(\Leftrightarrow7^{2x}=7^2\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
MT
1
IM
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 \(\left(x;y\right)=\left(\frac{1}{2};-1\right)\)
