bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

\(\Leftrightarrow\frac{\left(x+a\right)\left(3a^2x^2+a^2+8ax+x^2+3\right)\left(3a^6x^6+27a^6x^4+33a^6x2+a^6+72a^5x^5+24a^5x^3+72a^{5x}+27a^4x^6+459a^4x^4+441a^4x^2+33a^4+240a^3x^5+800a^3x^3+240a^3x+33a^2x^6+441a^2x^4+459x^2a^2+27a^2+75ax^5+240ax^3+72ax+x^6+33x^4+27x^2+3\right)}{\left(a^2+3\right)\left(a^6+33a^4+27a^{2+3}\right)\left(x^{2+3}\right)\left(x^6+33x^4+27x^2+3\right)}=0\)

mấy nhân tử sau ko cần chú ý đâu :)) chỉ cần chú ý đến x+a=0 <=>x=-a thôi :)) 

bài này đúng 100% nhé chỉ sợ gõ sai thôi, ko tin có thể dùng máy tính kiểm tra 

\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)+9=0\)

\(\Leftrightarrow\left(x^2-3x+4\right)\left(x^2+3x-10\right)+9=0\)

\(\Leftrightarrow\left(x^2+3x-7+3\right)\left(x^2+3x-7-3\right)+9=0\)

\(x^2+3x-7=0\)

\(x^2+3x=7\)

\(\Rightarrow x^2+2x.\frac{3}{2}+\frac{9}{4}=7+\frac{9}{4}\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2=\frac{37}{4}\)

\(\Rightarrow x+\frac{3}{2}=\pm\sqrt{\frac{37}{4}}\)

\(\Rightarrow x=\frac{-3}{2}-\sqrt{\frac{37}{4}}\)

\(\Rightarrow x=\frac{-3}{2}+\sqrt{\frac{37}{4}}\)

Vậy \(S=\left\{\frac{-3}{2}-\sqrt{\frac{37}{4}};\frac{-3}{2}+\sqrt{\frac{37}{4}}\right\}\)

\(4.\left(x+1\right)^2-9.\left(x-1\right)^2=0\)

\(\Leftrightarrow4.\left(x^2+2x+1\right)-9.\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow4x^2+8x+4-9x^2+18x-9=0\)

\(\Leftrightarrow\left(4x^2+8x+4\right)-\left(9x^2-18x+9\right)=0\)

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

\(\Leftrightarrow\left[2x+2-\left(3x-3\right)\right].\left[2x+2+\left(3x-3\right)\right]=0\)

\(\Leftrightarrow\left(2x+2-3x+3\right).\left(2x+2+3x-3\right)=0\)

\(\Leftrightarrow\left(5-x\right).\left(5x-1\right)=0\)

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{5;\frac{1}{5}\right\}.\)

Chúc bạn học tốt!

\(4\left(x+1\right)^2-9\left(x-1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+2x+1\right)-9\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow4x^2+8x+4-9x^2-18x-9=0\)

\(\Leftrightarrow-5x^2-10x-5=0\)

\(\Leftrightarrow-5\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow-5\left(x+1\right)^2=0\)

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy S = {1}

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

\(\frac{-5}{9}x+1=\frac{2}{3}x-10\)

\(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\)

\(-5x+9=6x-90\)

\(-5x-6x=-90-9\)

\(-11x=-99\)

\(x=\frac{-99}{-11}=9\)

b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)

\(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)

\(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)

\(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)

x=30

Chúc bạn học tốt!!