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3x2 + 2x - 1 = 0
=> 3x2 + 3x - x - 1 = 0
=> 3x(x + 1) - (x + 1) = 0
=> (3x - 1)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
x2 - 5x + 6 = 0
=> x2 - 2x - 3x + 6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 3)(x - 2) = 0
=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=> \(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)
2, \(x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}}\)
\(3x^2-7x-4=0\)
\(\Leftrightarrow3x^2-3x+4x-4=0\)
\(\Leftrightarrow3x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=1\end{matrix}\right.\)
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
a.
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)+3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
f.
\(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-3\left(x-1\right)\right]\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\\x=\pm\sqrt{3}\end{matrix}\right.\)
Cái này t dùng máy tính
\(\left(x-2\right)\left(x+3\right)\left(2x+1\right)\left(3x-1\right)=0\)
Đến đây thì pt có 4 nghiệm:\(x=2;-3;-\frac{1}{2};\frac{1}{3}\)
Vậy....
Ta có : \(\left(36x^2+84x+49\right)\left(3x^2+7x+4\right)=6\)
\(\Leftrightarrow\left(36x^2+84x+49\right)\left(36x^2+84x+48\right)=72\)(2)
Đặt : \(36x^2+84x+49=a\) Khi đó pt (2) có dạng :
\(a.\left(a-1\right)=72\)
\(\Leftrightarrow\left(a-9\right)\left(a+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=9\\a=-8\end{cases}}\)
+) Với \(a=9\Rightarrow36x^2+84x+49=9\)
\(\Leftrightarrow\left(6x+7\right)^2=9\)
\(\Leftrightarrow\orbr{\begin{cases}6x+7=3\\6x+7=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{cases}}\) ( thỏa mãn )
+) Với \(a=-8\Rightarrow36x^2+84x+49=-8\)
\(\Leftrightarrow\left(6x+7\right)^2=-8\) ( vô lí )
Vậy pt đã cho có tập nghiệm \(S=\left\{-\frac{2}{3},-\frac{5}{3}\right\}\)
Đặt \(u=3x^2+7x+4\)
Phương trình trở thành \(\left(12u+1\right)u=6\)
\(\Leftrightarrow12u^2+u-6=0\)
Ta có \(\Delta=1^2+4.12.6=289,\sqrt{\Delta}=17\)
\(\Rightarrow\orbr{\begin{cases}u=\frac{-1+17}{24}=\frac{2}{3}\\u=\frac{-1-17}{24}=\frac{-3}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x^2+7x+4=\frac{2}{3}\\3x^2+7x+4=\frac{-3}{4}\end{cases}}\)
+) \(3x^2+7x+4=\frac{2}{3}\)
\(\Rightarrow3x^2+7x+\frac{10}{3}=0\)
Ta có \(\Delta=7^2-4.3.\frac{10}{3}=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-7+3}{6}=\frac{-2}{3}\\x=\frac{-7-3}{6}=\frac{-5}{3}\end{cases}}\)
+) \(3x^2+7x+4=\frac{-3}{4}\)
\(\Rightarrow3x^2+7x+\frac{19}{4}=0\)
Ta có \(\Delta=7^2-4.3.\frac{19}{4}=-8< 0\)(vô nghiệm)
Tóm lại, phương trình chỉ có 2 nghiệm \(\left\{\frac{-2}{3};\frac{-5}{3}\right\}\)