Verified answer

Відповідь:

t ∈ (1, 2]

Пояснення:

[tex]\frac{\sqrt{3-t}+4t-7}{t-1}\leq2[/tex]

[tex]\frac{\sqrt{3-t}+4t-7}{t-1}-2\leq0[/tex]

[tex]\frac{\sqrt{3-t}+4t-7-2(t-1)}{t-1}\leq0[/tex]

[tex]\frac{\sqrt{3-t}+2t-5}{t-1}\leq0[/tex]

t ∈ (1,2]; t ∈ (-∞, 1); t ∈ (1,3]  ⇒  t ∈ (1,2]

[tex]\sqrt{3-t}+4t-7=t-1[/tex]

3-t=36-36t+9t²

9t²-35t-33=0

[tex]t=\frac{-(-35)+-\sqrt{(-35)^2-4*9*33}}{2*9}[/tex]

[tex]t=\frac{35+-\sqrt{1225-1188}}{18}[/tex]

{[tex]t=\frac{35+-\sqrt{37}}{18}[/tex]