\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\) \(\left(\dfrac{3}{2}\text{≤}x\text{≤}\dfrac{5}{2}\right)\)

\(\text{⇔}2\sqrt{2x-3}+2\sqrt{5-2x}=6x^2-24x+28\)

\(\text{⇔}2x-3-2\sqrt{2x-3}+1+5-2x-2\sqrt{5-2x}+1+6x^2-24x+24=0\) \(\text{⇔}\left(\sqrt{2x-3}-1\right)^2+\left(\sqrt{5-2x}-1\right)^2+6\left(x-2\right)^2=0\)

\(\text{⇔}x=2\left(TM\right)\)

KL.............

VT = \(\text{ }\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=\sqrt{2.2}=2\)

Dấu bằng xảy ra khi 2x - 3 = 5- 2x => x  = 2 

VP = \(3x^2-12x+14=3\left(x^2-4x+4\right)+2=3\left(x-2\right)^2+2\ge2\)

Dấu = xảy ra khi x = 2 

=> VT = VP = 2 khi x = 2 

a.ĐKXĐ:\(\frac{3}{2}\le x\le\frac{5}{2}\)

AD BĐT Cauchy ta được:

\(\sqrt{\left(2x-3\right)1}\le\frac{2x-3+1}{2}=\frac{2x-2}{2}=x-1\)

\(\sqrt{\left(5-2x\right)\cdot1}\le\frac{5-2x+1}{2}=\frac{6-2x}{2}=3-x\)

Do đó \(\sqrt{2x-3}+\sqrt{5-2x}\le x-1+3-x=2\)(1)

Lại có \(3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)(2)

Từ (1) và (2) suy ra \(\sqrt{2x-3}+\sqrt{5-2x}\le3x^2-12x+14\)

Dấu = khi x=2 (tm ĐKXĐ)

PHẦN b giải tương tự

ĐKXĐ: ...

\(VT\le\sqrt{2\left(2x-3+5-2x\right)}=2\)

\(VP=3\left(x-2\right)^2+2\ge2\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}2x-3=5-2x\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow x=2\)

1.

ĐKXĐ: ...

\(x^2-x+2=1\sqrt{x^2+x-1}+1\sqrt{x-x^2+1}\)

\(\Rightarrow x^2-x+2\le\dfrac{1}{2}\left(1+x^2+x-1\right)+\dfrac{1}{2}\left(1+x-x^2+1\right)\)

\(\Rightarrow x^2-2x+1\le0\)

\(\Rightarrow\left(x-1\right)^2\le0\)

\(\Rightarrow x=1\)

Thử lại ta thấy thỏa mãn

b.

ĐKXĐ: ...

Ta có:

\(VP=3\left(x-2\right)^2+2\ge2\)

\(VT=1\sqrt{2x-3}+1\sqrt{5-2x}\le\dfrac{1}{2}\left(1+2x-3\right)+\dfrac{1}{2}\left(1+5-2x\right)=2\)

\(\Rightarrow VT\le VP\)

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}x-2=0\\1=2x-3\\1=5-2x\end{matrix}\right.\) \(\Leftrightarrow x=2\)

\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\left(1\right)\)

ĐKXĐ: \(\frac{3}{2}\le x\le\frac{5}{2}\)

Áp dụng BĐT Bunhiacopxki ta có:

\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)

Dấu "=" xảy ra <=> \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)

Ta lại có VP=3x²-12x+14=3(x-2)²+2 >=2

Dấu "=" xảy ra khi x=2

Do đó VT=VP <=> x=2 (ttmđk)

Vậy S={2}

Ta có:

\(\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{\left(1+1\right)\left(2x-3+5-2x\right)}=2\)

\(3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)

\(\Rightarrow\) phương trình có nghiệm khi và chỉ khi:

\(\left\{{}\begin{matrix}3\left(x-2\right)^2=0\\2x-3=5-2x\end{matrix}\right.\) \(\Rightarrow x=2\)

TXĐ : \(\left[\dfrac{3}{2};\dfrac{5}{2}\right]\).

Áp dụng BĐT Bunhiacopski :

\(VT^2=\left(\sqrt{2x-3}+\sqrt{5-2x}\right)^2\le\left(1^2+1^2\right)\left(2x-3+5-2x\right)=4\)

\(\Rightarrow VT\le2\)

Xảy ra khi \(\dfrac{\sqrt{2x-3}}{1}=\dfrac{\sqrt{5-2x}}{1}\Rightarrow x=2\)(1)

\(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)

Xảy ra khi \(x-2=0\Leftrightarrow x=2\)(2)

Từ (1)(2) => Pt có nghiệm x=2

đã làm