Mình fix luôn đề nhé.
Ta có :
+) \(\sqrt{3x^2-18x+28}=\sqrt{3\left(x^2-6x+9\right)+1}\)
\(=\sqrt{3\left(x-3\right)^2+1}\ge1\forall x\)
+) \(\sqrt{4x^2-24x+45}=\sqrt{4\left(x^2-6x+9\right)+9}\)
\(=\sqrt{3\left(x-3\right)^2+9}\ge\sqrt{9}=3\forall x\)
Do đó \(VT\ge4\forall x\)
Xét \(VP=-5-x^2+6x\)
\(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=4-\left(x-3\right)^2\le4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\).
\(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6x\)
<=> \(\sqrt{3\left(x^2-6x+9\right)+1}+\sqrt{4\left(x^2-6x+9\right)+9}=4-\left(x^2-6x+9\right)\)
<=> \(\sqrt{3\left(x-3\right)^2+1}+\sqrt{4\left(x-3\right)^2+9}=4-\left(x-3\right)^2\)
Có \(\sqrt{3\left(x-3\right)^2+1}\ge\sqrt{0+1}=1\)
\(\sqrt{4\left(x-3\right)^2+9}\ge\sqrt{0+9}=3\)
=> VT=\(\sqrt{3\left(x-3\right)^2+1}+\sqrt{4\left(x-3\right)^2+9}\ge1+3=4\)
VP=\(4-\left(x-3\right)^2\le4\) với mọi x
=> Để VT=VP <=> \(x-3=0\) <=>x=3(t/m)
Vậy pt có nghiệm x=3