\(\left(2x-1\right)\left(x+7\right)=x^2-49\)

\(\Leftrightarrow\left(2x-1\right)\left(x+7\right)=\left(x-7\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-6\end{matrix}\right.\)

a) \(ĐKXĐ:2x^2+6x+1\ge0\)

Với \(x\ge2\) pt cho trở thành :

\(2x^2+6x+1=x^2+4x+4\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=3\) ( do \(x\ge2\) )

Vậy pt có tập nghiệm \(S=\left\{3\right\}\)

c) \(\left|x^2-2x-3\right|+\left|x+1\right|=0\)

\(\Leftrightarrow\left|\left(x+1\right)\left(x-3\right)\right|+\left|x+1\right|=0\)

\(\Leftrightarrow\left|x+1\right|.\left(\left|x-3\right|+1\right)=0\)

\(\Leftrightarrow\left|x+1\right|=0\) ( Do \(\left|x-3\right|+1>0\) )

\(\Leftrightarrow x=-1\)

Vậy $x=-1$

\(A=x^2+x+5=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\)

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

Dấu "=" xảy ra khi \(x=\dfrac{3}{2}\)

Ta có : \(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)

\(\Leftrightarrow2\cdot\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{x}{98}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{98}\)

\(\Leftrightarrow\dfrac{2}{7}=\dfrac{x}{98}\Rightarrow x=28\)

Vậy $x=28$

\(ĐKXĐ:\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\5-4x^2\ge0\end{matrix}\right.\) (*)

Ta có pt cho tương đương :

\(8-6\sqrt{2x-1}-2x\sqrt{5-4x^2}=0\)

\(\Leftrightarrow\left(5-4x^2-2x\sqrt{5-4x^2}+x^2\right)+3.\left(2x-1-2\sqrt{2x-1}+1\right)+3\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{5-4x^2}-x\right)^2+3\left(\sqrt{2x-1}-1\right)^2+3\left(x-1\right)^2=0\)

Dấu "=" xảy ra khi và chỉ ra \(x=1\) ( Thỏa mãn (*) )

Vậy pt có nghiệm duy nhất \(x=1\)

18. B => quickly

19. B => which

20. A => If

\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1+\sqrt{7}+1+\sqrt{14}}{\sqrt{2}}\)

\(=\sqrt{14}+\sqrt{7}\)

\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)

\(=\sqrt{2}.\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)

\(=\sqrt{2}\cdot\left(\sqrt{5}-1+1+\sqrt{5}\right)=2\sqrt{10}\)

Bạn tham khảo lời giải mình bắt gặp được :

\(-\dfrac{5}{12}\cdot\dfrac{6}{7}+\dfrac{7}{12}\left(-\dfrac{3}{14}\right)\)

\(=\dfrac{1}{12}\cdot\left(\dfrac{-5\cdot6}{7}+\dfrac{7\cdot\left(-3\right)}{14}\right)\)

\(=\dfrac{1}{12}\cdot\left(-\dfrac{81}{14}\right)=-\dfrac{27}{56}\)