a) \(\Leftrightarrow\left(-63x^2+78x-15\right)+\left(63x^3+x-20\right)=44\)

\(\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\)

\(\Leftrightarrow79x-35=44\)

\(\Leftrightarrow79x=44+35\)

\(\Leftrightarrow79x=79\)

\(\Leftrightarrow x=1\)

b) \(\Leftrightarrow\left(x^2+3x+2\right).\left(x+5\right)-x^2.\left(x+8\right)=27\)

\(\Leftrightarrow x.\left(x^2+3x+2\right)+5.\left(x^2+3x+2\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x+10=27\)

\(\Leftrightarrow17x=17\)

\(\Leftrightarrow x=1\)

Điều kiện x khác 0

     \(\left(5x^4-3x^3\right):2x^3=\frac{1}{2}\) 

\(\Rightarrow\frac{5}{2}x-\frac{3}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{5}{2}x=2\Rightarrow x=\frac{4}{5}\)

Lần sau bạn chú ý viết đầy đủ đề.

1.

\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)

2.

\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)

Phạm Mạnh Kiên: sửa lại theo ý bạn thì làm như sau:

1.

\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{4}+4}-\sqrt{5-2\sqrt{5}.\sqrt{4}+4}\)

\(=\sqrt{(\sqrt{5}+\sqrt{4})^2}-\sqrt{(\sqrt{5}-\sqrt{4})^2}=|\sqrt{5}+2|-|\sqrt{5}-2|\)

\(=\sqrt{5}+2-(\sqrt{5}-2)=4\)

2.

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}=|\sqrt{7}-1|-|\sqrt{7}+1|\)

\(=-2\)

1) \(2x^4+3x^3-x^2+3x+2=0\)

\(\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0\)

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

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

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

\(\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0\)

\(\Rightarrow\left(2x+1\right)\left(x+2\right)\left(x^2-x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\x^2-x+1=0\end{matrix}\right.\)

Ta có:

\(x^2-x+1\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x

\(\Rightarrow x^2-x+1\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

3) \(\left(x+2\right)^4+\left(x+4\right)^4=16\)

Đặt x + 3 = a, ta được

\(\left(a-1\right)^4+\left(a+1\right)^4=16\)

\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16\)

\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)

\(\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16\)

\(\Rightarrow2a^4+2.4a^2+2+2.2a^2=16\)

\(\Rightarrow2a^4+8a^2+4a^2+2=16\)

\(\Rightarrow2a^4+12a^2+2-16=0\)

\(\Rightarrow2a^4+12a^2-14=0\)

\(\Rightarrow2a^4-2a^2+14a^2-14=0\)

\(\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0\)

\(\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+7=0\end{matrix}\right.\)

Vì \(a^2\ge0\) với mọi a

\(\Rightarrow a^2+7\ge7\) với mọi a

\(\Rightarrow a^2+7\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+3-1=0\\x+3+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

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

\(\Leftrightarrow x=6\)