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Giải Phương Trình Sau (Nhớ ghi cách làm nha mình k đúng cho)
bài 1+2: phân tích mẫu thành nhân tử r` áp dụng
1/ab=1/a-1/b
bài 3+4: quy đồng rút gọn blah...
Ta có: \(\frac{1}{x^2+9x+20}\)\(+\frac{1}{x^2+11x+30}\)\(+\frac{1}{x^2+13x+42}\)
=\(\frac{1}{x^2+4x+5x+20}\)\(+\frac{1}{x^2+5x+6x+30}\)\(+\frac{1}{x^2+6x+7x+42}\)
=\(\frac{1}{x\left(x+4\right)+5\left(x+4\right)}\)\(+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}\)\(+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}\)
=\(\frac{1}{\left(x+4\right)\left(x+5\right)}\)\(+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(+\frac{1}{\left(x+6\right)\left(x+7\right)}\)
=\(\frac{1}{x+4}-\frac{1}{x+5}\)\(+\frac{1}{x+5}-\frac{1}{x+6}\)\(+\frac{1}{x+6}-\frac{1}{x+7}\)
=\(\frac{1}{x+4}-\frac{1}{x+7}\)
=\(\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}\)=\(\frac{3}{\left(x+4\right)\left(x+7\right)}\)
\(\frac{1}{x^2+9x+20}\) \(+\) \(\frac{1}{x^2+11x+30}\) \(+\)\(\frac{1}{x^2+13x+42}\)
= \(\frac{1}{\left(x+4\right)\left(x+5\right)}\)\(+\) \(\frac{1}{\left(x+5\right)\left(x+6\right)}\) \(+\) \(\frac{1}{\left(x+6\right)\left(x+7\right)}\)
= \(\frac{1}{x+4}\)\(-\)\(\frac{1}{x+5}\) \(+\)\(\frac{1}{x+5}\)\(-\)\(\frac{1}{x+6}\)\(+\)\(\frac{1}{x+6}\)\(-\)\(\frac{1}{x+7}\)
= \(\frac{1}{x+4}\)\(-\)\(\frac{1}{x+7}\)= \(\frac{x+7-\left(x+4\right)}{\left(x+4\right)\left(x+7\right)}\)= \(\frac{3}{x^2+11x+28}\)
\(A=\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\Leftrightarrow\frac{1-x+3x+3-2x-3}{x+1}=\frac{1}{x+1}=0\)
Vô nghiệm.
\(B=\left(5,5-11x\right)\left(\frac{7x+2}{5}+\frac{2\left(1-3x\right)}{3}\right)=0\)
\(\left[\begin{matrix}5,5-11x=0\\3\left(7x+2\right)+10-30x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{5,5}{11}\\-9x+16=0\end{matrix}\right.\)\(\left[\begin{matrix}x=\frac{1}{2}\\x=\frac{16}{9}\end{matrix}\right.\)
b) (5,5-11x)(\(\frac{7x+2}{5}\)+ \(\frac{2\left(1-3x\right)}{3}\)) = 0
<=> (5,5 - 11x )(\(\frac{-9x+16}{15}\))=0
<=>\(\left[\begin{matrix}5,5-11x=0\\\frac{-9x+16}{15}=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=\frac{16}{9}\end{matrix}\right.\)
Vậy pt có nghiệm là x=\(\frac{1}{2}\) và x= \(\frac{16}{9}\)
b, (3,3-11x)(\(\frac{7x+2}{5}\)+\(\frac{2\left(1-3x\right)}{3}\))=0
Câu c : \(x^4-3x^3+2x^2-9x+9=0\)
<=>\(x^4-x^3-2x^3+2x^2-9x+9=0\)
<=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
<=> \(x-1=0\) hoặc \(x^3-2x^2-9=0\)
Nếu x-1=0 <=> x=1
Nếu \(x^3-2x^2-9=0\)
<=> \(x^3-3x^2+x^2-9=0\)
<=>\(x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
<=>\(\left(x-3\right)\left(x^2+x+3\right)=0\)
Vì \(x^2+x+3=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\) >0 nên x-3=0 <=> x=3
Vậy \(S=\left\{1;3\right\}\)
Câu b : \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
<=> \(4x^2\left(x^2+2x+2\right)=5\left(x^2+2x+1\right)\)
<=> \(4x^4+8x^3+8x^2=5x^2+10x+5\)
<=>\(4x^4+8x^3+3x^2-10x-5=0\)
<=>\(4x^4-4x^3+12x^3-12x^2+15x^2-15x+5x-5=0\)
<=>\(\left(x-1\right)\left(4x^3+12x^2+15x+5\right)=0\)
<=>\(\left(x-1\right)\left(2x+1\right)\left(2x^2+5x+5\right)=0\)
<=>x=1 hoặc \(x=\frac{-1}{2}\)
Phương trình \(2x^2+5x+5=0\) Vô nghiệm
Đặt
6x+7 = 7 , ta có
\(\left(t+1\right)\left(t-1\right)t^2=72\Rightarrow\left(t^2-1\right)t^2=72\)
\(\Rightarrow t^4-t^2-72=0\)
Lại đặt \(t^2=a\) (a \(\ge0\) )
\(\Rightarrow a^2-a-72=0\Rightarrow\left(a+8\right)\left(a-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-8\left(ktm\right)\\a=9\left(tm\right)\end{matrix}\right.\)
a = 9 => \(\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\)
Với t = 3
=> 6x + 7 =3
=> 6x = -4
=> x= \(-\frac{2}{3}\)
Với t = -3
=> 6x + 7 = -3
=> 6x = -10
=> x = \(-\frac{5}{3}\)
Vậy.....
b)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x-4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\Rightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+7\right)\left(x+4\right)}=\frac{1}{18}\Rightarrow x^2+11x+28-54=0\Rightarrow x^2+11x-26=0\)
\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
a) Ta có:
(6x+8)(6x+6)(6x+7)2 = 72
Đặt \(6x+7=a\)
\(\Rightarrow\left(a+1\right)\left(a-1\right)a^2=72\)
\(\Leftrightarrow a^4-a^2-72=0\)
\(\Leftrightarrow\left(a^4+8a^2\right)+\left(-9a^2-72\right)=0\)
\(\Leftrightarrow\left(a^2+8\right)\left(a^2-9\right)=0\)
Đễ thấy \(a^2+8>0\)
\(\Rightarrow a^2-9=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x+7=3\\6x+7=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)
b)
a) Ta có: \(\frac{2\left(x-4\right)}{3}+\frac{3x+13}{8}=\frac{2\left(2x-3\right)}{5}+12\)
⇔\(\frac{80\left(x-4\right)}{120}+\frac{15\left(3x+13\right)}{120}-\frac{48\left(2x-3\right)}{120}-\frac{1440}{120}=0\)
⇔\(80\left(x-4\right)+15\left(3x+13\right)-48\left(2x-3\right)-1440=0\)
\(\Leftrightarrow80x-320+45x+195-96x+144-1440=0\)
⇔\(29x-1421=0\)
\(\Leftrightarrow29x=1421\)
hay x=49
Vậy: x=49
b) Ta có: \(\frac{2\left(5x+2\right)}{9}-1=\frac{4\left(33+2x\right)}{5}-\frac{5\left(1-11x\right)}{9}\)
⇔\(\frac{10\left(5x+2\right)}{45}-\frac{45}{45}-\frac{36\left(33+2x\right)}{45}+\frac{25\left(1-11x\right)}{45}=0\)
⇔\(10\left(5x+2\right)-45-36\left(33+2x\right)+25\left(1-11x\right)=0\)
\(\Leftrightarrow50x+20-45-1188-72x+25-275x=0\)
\(\Leftrightarrow-297x-1188=0\)
\(\Leftrightarrow-297x=1188\)
hay x=-4
Vậy: x=-4