tất cả đều mũ chẳn nên lớn hơn hoặc bằng 0 => để thõa mãn các tổng cộng lại bằng 0 => mỗi tổng bằng 0 

a, Vì \(\hept{\begin{cases}\left(12a-9\right)^2\ge0\\\left(8b+1\right)^4\ge0\\\left(c+15\right)^6\ge0\end{cases}\Rightarrow\left(12a-9\right)^2+\left(8b+1\right)^4+\left(c+15\right)^6\ge0}\)

Mà \(\left(12a-9\right)^2+\left(8b+1\right)^4+\left(c+15\right)^6\le0\)

\(\Rightarrow\hept{\begin{cases}\left(12a-9\right)^2=0\\\left(8b+1\right)^4=0\\\left(c+15\right)^6=0\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{3}{4}\\b=\frac{-1}{8}\\c=-15\end{cases}}}\)

b, tương tự a

\(\Rightarrow-3< x< 2\\ \Rightarrow x\in\left\{-2;-1;0;1\right\}\\ \Rightarrow B\)

B     nhé

>_<

a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)

Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)

Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)

- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)

Vậy ...

b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;

ok

A

Chọn A

1) \(-4< x< 3\)

\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2\right\}\)

Tổng:

\(\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2\)

\(=\left(-2+2\right)+\left(-1+1\right)+0-3\)

\(=-3\)

2) \(-5< x< 5\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

Tổng:

\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+3\)

\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0\)

\(=0\)

3) \(-10< x< 6\)

\(\Rightarrow x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

Tổng:

\(\left(-9\right)+\left(-8\right)+\left(-7\right)++\left(-6\right)+\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4+5\)

\(=-24\)

4) \(-6< x< 5\)

\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4\right\}\)

Tổng:

\(\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4\)

\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0-5\)

\(=-5\)

5) \(-5< x< 2\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1\right\}\)

Tổng:

\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1\)

\(=\left(-1+1\right)+0+\left(-4-3-2\right)\)

\(=-6\)

c: TH1: m=1

Phương trình sẽ trở thành:

\(\left(1-1\right)x^2-2\cdot1\cdot x+1+1=0\)

=>-2x+2=0

=>-2x=-2

=>x=1

=>Loại

TH2: m<>1

\(\Delta=\left(-2m\right)^2-4\left(m-1\right)\left(m+1\right)=4m^2-4\cdot\left(m^2-1\right)=4m^2-4m^2+4=4\) >0

=>Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{2m-\sqrt4}{2\cdot\left(m-1\right)}=\frac{2m-2}{2\left(m-1\right)}=1\\ x=\frac{2m+\sqrt4}{2\left(m-1\right)}=\frac{2m+2}{2\left(m-1\right)}=\frac{m+1}{m-1}\end{array}\right.\)

\(4\left(x_1^2+x_2^2\right)=5x_1^2\cdot x_2^2\)

=>\(4\left\lbrack1^2+\left(\frac{m+1}{m-1}\right)^2\right\rbrack=5\cdot1^2\cdot\left(\frac{m+1}{m-1}\right)^2\)

=>\(5\left(\frac{m+1}{m-1}\right)^2=4\left(\frac{m+1}{m-1}\right)^2+4\)

=>\(\left(\frac{m+1}{m-1}\right)^2=4\)

=>\(\left[\begin{array}{l}\frac{m+1}{m-1}=2\\ \frac{m+1}{m-1}=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}2\left(m-1\right)=m+1\\ -2\left(m-1\right)=m+1\end{array}\right.\)

=>\(\left[\begin{array}{l}2m-2=m+1\\ -2m+2=m+1\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\left(nhận\right)\\ -3m=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\left(nhận\right)\\ m=\frac13\left(nhận\right)\end{array}\right.\)

a: TH1: m=-1

Phương trình sẽ trở thành:

\(\left(-1+1\right)x^2-2\left(-1+1\right)x+\left(-1\right)-3=0\)

=>-4=0(vô lý)

TH2: m<>-1

\(\Delta=\left\lbrack-2\left(m+1\right)\right\rbrack^2-4\left(m+1\right)\left(m-3\right)\)

\(=4\left(m^2+2m+1\right)-4\left(m^2-2m-3\right)\)

\(=4m^2+8m+4-4m^2+8m+12=16m+16\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>16m+16>0

=>m+1>0

=>m>-1

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=\frac{2\left(m+1\right)}{m+1}=2;x_1x_2=\frac{c}{a}=\frac{m-3}{m+1}\)

\(\left(4x_1+1\right)\left(4x_2+1\right)=18\)

=>\(16x_1x_2+4\left(x_1+x_2\right)+1=18\)

=>\(16\cdot\frac{m-3}{m+1}+4\cdot2+1=18\)

=>\(16\cdot\frac{m-3}{m+1}=18-8-1=18-9=9\)

=>\(\frac{m-3}{m+1}=\frac{9}{16}\)

=>16(m-3)=9(m+1)

=>16m-48=9m+9

=>7m=57

=>m=57/7(nhận)

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