Ta có

M ( x ) = P ( x ) − Q ( x ) = − 6 x 5 − 4 x 4 + 3 x 2 − 2 x − 2 x 5 − 4 x 4 − 2 x 3 + 2 x 2 − x − 3 = − 8 x 5 + 2 x 3 + x 2 − x + 3  Có  M ( − 1 ) = − 8. ( − 1 ) 5 + 2 ⋅ ( − 1 ) 3 + ( − 1 ) 2 − ( − 1 ) + 3 = 11

Chọn đáp án A

Đáp án  B

4

Chọn A.

ĐK: 1 ≤ x ≤ 7

Ta có 

Xét y(1) = y(7) = 6 , y(4) =  2 3 suy ra 2,44 < k < 3,464  suy ra k = 3 có 1 số nguyên k.

Trả lời đc tui chết liền

Sorry

Bài 2:
\(41-2^x-1=9\)

\(\Rightarrow40-2^x=9\)

\(\Rightarrow2^x=31\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)

Vậy không có giá trị x thỏa mãn

b) \(\left|x-3\right|-21=-5\)

\(\Rightarrow\left|x-3\right|=16\)

\(\Rightarrow x-3=\pm16\)

+) \(x-3=16\Rightarrow x=19\)

+) \(x-3=-16\Rightarrow x=-13\)

Vậy \(x\in\left\{19;-13\right\}\)

d) Ta có: \(M=7+7^2+...+7^{2016}\)

\(\Rightarrow7M=7^2+7^3+...+7^{2017}\)

\(\Rightarrow7M-M=\left(7^2+7^3+...+7^{2017}\right)-\left(7+7^2+...+7^{2016}\right)\)

\(\Rightarrow6M=7^{2017}-7\)

\(6M+7=7^{x+1}\)

\(\Rightarrow7^{2017}-7+7=7^{x-1}\)

\(\Rightarrow7^{x-1}=7^{2017}\)

\(\Rightarrow x-1=2017\)

\(\Rightarrow x=2018\)

Vậy \(x=2018\)

g) Ta có: \(x+9⋮x-2\)

\(\Rightarrow\left(x-2\right)+11⋮x-2\)

\(\Rightarrow11⋮x-2\)

\(\Rightarrow x-2\in\left\{1;-1;11;-11\right\}\)

+) \(x-2=1\Rightarrow x=3\)

+) \(x-2=-1\Rightarrow x=1\)

+) \(x-2=11\Rightarrow x=13\)

+) \(x-2=-11\Rightarrow x=-9\)

Vậy \(x\in\left\{3;1;13;-9\right\}\)

Bài 3:

a) \(\left(x+1\right)^{100}+\left|y-3\right|=0\)

\(\Rightarrow\left(x+1\right)^{100}=0\) và \(\left|y-3\right|=0\)

+) \(\left(x+1\right)^{100}=0\Rightarrow x+1=0\Rightarrow x=-1\)

+) \(\left|y-3\right|=0\Rightarrow y-3=0\Rightarrow y=3\)

Vậy \(x=-1;y=3\)

Giúp mình với

Chọn đáp án A

\(\Delta=\left(2m+3\right)^2-4m=4m^2+12m+9-4m=4m^2+8m+9\)

\(=4\left(m^2+2m+1-1\right)+9=4\left(m+1\right)^2+5\ge5>0\forall m\)

Vậy pt luôn có 2 nghiệm pb 

\(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=m\end{matrix}\right.\)Ta có : \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(\left(2m+3\right)^2-2m=4m^2+12m+9-2m=4m^2+10m+9\)

\(=4m^2+\dfrac{2.2m.10}{4}+\dfrac{100}{16}-\dfrac{100}{16}+9\)

\(=\left(2m+\dfrac{10}{4}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall m\)

Dấu ''='' xảy ra khi x = -5/4 

Đáp án C.