\(\Delta=\left\lbrack2\left(m+1\right)\right\rbrack^2-4\cdot1\cdot\left(2m-3\right)\) \(=4\left(m^2+2m+1\right)-4\left(2m-3\right)=4\left(m^2+2m+1-2m+3\right)=4\left(m^2+4\right)\) >0=>Phương trình luôn có hai nghiệm phân biệtTHeo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m+1\right);x_1x_2=\frac{c}{a}=2m-3\) \(x_1^2+x_2^2\) \(=\left(x_1+x_2\right)^2-2x_1x_2\) \(=\left(2m+2\right)^2-2\left(2m-3\right)=4m^2+8m+4-4m+6=4m^2+4m+10\) \(\left(x_1-1\right)\left(x_2-1\right)\) =x1x2-(x1+x2)+1=2m-3-(2m+2)+1=2m-3-2m-2+1=-3-2+1=-5+1=-4\(\left|x_1-1\right|+\left|x_2-1\right|=4\) =>\(\left(\left|x_1-1\right|+\left|x_2-1\right|\right)^2=16\) =>\(\left(x_1-1\right)^2+\left(x_2-1\right)^2+2\cdot\left|\left(x_1-1_{}\right)\left(x_2-1\right)\right|=16\) =>\(\left(x_1^2+x_2^2\right)-2\left(x_1+x_2\right)+2+2\cdot\left|-4\right|=16\) =>\(4m^2+4m+10\) -2(2m+2)+2+8=16=>\(4m^2+4m+20-4m-4=16\) =>\(4m^2=0\) =>m=0Câu trả lời: \(\Delta=\left\lbrack2\left(m+1\right)\right\rbrack^2-4\cdot1\cdot\left(2m-3\right)\) \(=4\left(m^2+2m+1\right)-4\left(2m-3\right)=4\left(m^2+2m+1-2m+3\right)=4\left(m^2+4\right)\) >0=>Phương trình luôn có hai nghiệm phân biệtTHeo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m+1\right);x_1x_2=\frac{c}{a}=2m-3\) \(x_1^2+x_2^2\) \(=\left(x_1+x_2\right)^2-2x_1x_2\) \(=\left(2m+2\right)^2-2\left(2m-3\right)=4m^2+8m+4-4m+6=4m^2+4m+10\) \(\left(x_1-1\right)\left(x_2-1\right)\) =x1x2-(x1+x2)+1=2m-3-(2m+2)+1=2m-3-2m-2+1=-3-2+1=-5+1=-4\(\left|x_1-1\right|+\left|x_2-1\right|=4\) =>\(\left(\left|x_1-1\right|+\left|x_2-1\right|\right)^2=16\) =>\(\left(x_1-1\right)^2+\left(x_2-1\right)^2+2\cdot\left|\left(x_1-1_{}\right)\left(x_2-1\right)\right|=16\) =>\(\left(x_1^2+x_2^2\right)-2\left(x_1+x_2\right)+2+2\cdot\left|-4\right|=16\) =>\(4m^2+4m+10\) -2(2m+2)+2+8=16=>\(4m^2+4m+20-4m-4=16\) =>\(4m^2=0\) =>m=0