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^A =130
^B =130
mà 2 góc này ở vị trí sole trong
⇒a//y(1)
^O=180⇒^O1=180-130
⇒^O1=50
Mà Ox là tia phân giác ⇒ ^O1=^O2=50
⇒^O2=B=50
Mà 2 góc này ở vị tri sole trong
⇒y//b(2)
Từ (1)(2)⇒a//b
\(P=x^4+2x^2+1+x^4+2x^3+x^2-3\)
\(P=2x^4+2x^3+3x^2-2\)
\(P=x^2\left(2x^2+2x+3\right)-2=\dfrac{1}{2}x^2\left[\left(2x+1\right)^2+5\right]-2\ge-2\)
\(P_{min}=-2\) khi \(x=0\)
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}\) đề ntn phải ko ạ
mà x+y+z=20
áp dụng DTSBN ta có
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{x+y+z}{5+2+3}=\dfrac{20}{10}=2\\ =>x=2\cdot5=10\\ y=2\cdot2=4\\ z=2\cdot3=6\)
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
\(24-16(x-\frac{1}{2})=23\)
\(16(x-\frac{1}{2})=24-23\)
\(16(x-\frac{1}{2})=1\)
\(x-\frac{1}{2}=\frac{1}{16}\)
\(x=\frac{1}{16}+\frac{1}{2}\)
\(x=\frac{9}{16}\)
Vậy số thực x cần tìm là \(\frac{9}{16}\)
Chúc bạn hok tốt ~
Bài 7:
Ta có: f(x)=4
\(\Leftrightarrow x-4-3\left(x+1\right)=4\)
\(\Leftrightarrow x-4-3x-3-4=0\)
\(\Leftrightarrow-2x=11\)
hay \(x=-\dfrac{11}{2}\)
saiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
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)
a) Ta có : \(\left|2x-1\right|\ge0\forall x\)
=> 5 + |2x - 1| \(\ge\)5 \(\forall x\)
=> Min A = 5
Dấu "=" xảy ra <=> 2x - 1 = 0 <=> x = 1/2
Vậy Min A = 5 <=> x = 1/2
b) Ta có \(\left|3y+12\right|\ge0\forall y\)
=> \(-\left|3y+12\right|\le0\forall y\)
=> \(4-\left|3y+12\right|\le4\forall y\)
=> Max B = 4
Dấu "=" xảy ra <=> 3y + 12 = 0
<=> y = -4
Vậy Max B = 4 <=> y = -4
thank XYZ