Làm cho em bài 26->35 đi ạ
View attachment 17989
26. Đặt $\sqrt{x^2+2451}=y \ (y\geq \sqrt{2451})$
$\Rightarrow (y^2-2451)^2+y=2451$
$\Leftrightarrow y^4-4902y^2+y+6004950=0$
$\Leftrightarrow (y-50)(y+49)(y^2+y-2451)=0$
$\Rightarrow y=50$ (vì $y\geq \sqrt{2451}$)
$\Rightarrow x^2+2451=2500\Leftrightarrow x=7$
27. VT $=|x|+|x-1|=|x|+|1-x|\geq |x+1-x|=1$
Dấu '=' xảy ra khi $x(1-x)\geq 0\Leftrightarrow 0\leq x\leq 1$
$\Rightarrow$ VT $=$ VP $\Leftrightarrow 0\leq x\leq 1$
28. VT $=|x-2|+|x+3|=|2-x|+|x+3|\geq |2-x+x+3|=5>$ VP
=> pt vô nghiệm
29. ĐK: $x\geq 1$
VT $=|\sqrt{x-1}-2|+|\sqrt{x-1}+3|=|2-\sqrt{x-1}|+|\sqrt{x-1}+3|\geq |2-\sqrt{x-1}+\sqrt{x-1}+3=5$
Dấu '=' xảy ra khi $1\leq x\leq 5$
............................
30. VT $=\sqrt{(x-2)^2+1}+\sqrt{(x-2)^2+4}+\sqrt{(x-2)^2+5}\geq \sqrt 1+\sqrt 4+\sqrt 5=\sqrt 5+3$
Dấu '=' xảy ra khi $x=2$
............................
31. VT $=\sqrt{2(x+1)^2+4}+\sqrt{(x^2-1)^2+9}\geq \sqrt 4+\sqrt 9=5$
VP $=4-2x-x^2=5-(x+1)^2\leq 5$
$\Rightarrow$ VT $=$ VP $\Leftrightarrow x=-1$
32. ĐK: $1\leq x\leq 9$
$(\sqrt{x-1}+\sqrt{9-x})^2\leq 2(x-1+9-x)=16\Rightarrow$ VT $\leq 4$
Dấu '=' xảy ra khi $x=5$
Mà VP $=x^2-10x+29=(x-5)^2+4\geq 4$. Dấu '=' xảy ra khi $x=5$
$\Rightarrow$ VT $=$ VP $\Leftrightarrow x=5$ (TM)
33. ĐK: $2\leq x\leq 10$
$(\sqrt{x-2}+\sqrt{10-x})^2\leq 2(x-2+10-x)=16\Rightarrow$ VT $\leq 4$
Dấu '=' xảy ra khi $x=6$
Mà VP $=x^2-12x+40=(x-6)^2+4\geq 4$. Dấu '=' xảy ra khi $x=56$
$\Rightarrow$ VT $=$ VP $\Leftrightarrow x=6$ (TM)
34. ĐK: $\dfrac 32\leq x\leq \dfrac 52$
$(\sqrt{2x-3}+\sqrt{5-2x})^2\leq 2(2x-3+5-2x)=4\Rightarrow$ VT $\leq 2$
Dấu '=' xảy ra khi $x=2$
Mà VP $=3x^2-12x+14=3(x-2)^2+2\geq 2$. Dấu '=' xảy ra khi $x=2$
$\Rightarrow$ VT $=$ VP $\Leftrightarrow x=2$ (TM)
35. ĐK: $\dfrac 53\leq x\leq \dfrac 73$
$(\sqrt{3x-5}+\sqrt{7-3x})^2\leq 2(3x-5+7-3x)=4\Rightarrow$ VT $\leq 2$
Dấu '=' xảy ra khi $x=2$
Mà VP $=5x^2-20x+22=5(x-2)^2+2\geq 2$. Dấu '=' xảy ra khi $x=2$
$\Rightarrow$ VT $=$ VP $\Leftrightarrow x=2$ (TM)