x²-4x+7 = 0 ⇔ x² -4x + 4 + 3 = 0 

⇔ (x-2)²+3=0 ⇔ (x-2)²=-3 (vô lí)

Vậy pt vô nghiệm

*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm

Ta có: \(x^2-4x+7=0\)

\(\Leftrightarrow x^2-4x+4+3=0\)

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

mà \(\left(x-2\right)^2+3\ge3>0\forall x\)

nên \(x\in\varnothing\)(đpcm)

b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)

\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)

\(\Leftrightarrow3x\left(x+4\right)=0\)

=>x=0(nhận) hoặc x=-4(loại)

Chọn B

Đáp án là B vì 12: -3 = -4; 12: -4 = -3; 12: -6 = -2;12: -12 = -1 và đáp ứng điều kiện a< -2

chắc là đúng đó

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bạn làm sai rồi !

\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)

\(\Leftrightarrow4x+26=-12\)

\(\Leftrightarrow4x=-38\)

\(\Leftrightarrow x=-\frac{19}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)

SAI GẦN HẾT

a/ \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)

\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)

Vậy..

b/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)

\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)

\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)

\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)

Vậy..

\(B=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(\Leftrightarrow B=\left(x^2+x+\dfrac{3}{2}-\dfrac{1}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{1}{2}\right)-12\)

\(\Leftrightarrow B=\left(x^2+x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}-12\)

\(\Leftrightarrow B=\left(x^2+x+\dfrac{3}{2}\right)^2-\dfrac{49}{4}\)

\(\Leftrightarrow B=\left(x^2+x+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{7}{2}\right)\)

\(\Leftrightarrow B=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(C=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(\Leftrightarrow C=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)

\(\Leftrightarrow C=\left(x^2+3x\right)\left(x^2+x+2x+2\right)+1\)

\(\Leftrightarrow C=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

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

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

\(\Leftrightarrow C=\left(x^2+3x+1\right)^2\)

\(D=\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(\Leftrightarrow D=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(\Leftrightarrow D=\left(x^2-x-7x+7\right)\left(x^2-3x-5x+15\right)-20\)

\(\Leftrightarrow D=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

\(\Leftrightarrow D=\left(x^2-8x+11-4\right)\left(x^2-8x+11+4\right)-20\)

\(\Leftrightarrow D=\left(x^2-8x+11\right)^2-16-20\)

\(\Leftrightarrow D=\left(x^2-8x+11\right)^2-36\)

\(\Leftrightarrow D=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)\)

\(\Leftrightarrow D=\left(x^2-8x+5\right)\left(x^2-8x+17\right)\)

:D