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a. x2 - 11 = \(x^2-\left(\sqrt{11}\right)^2=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)
b. x2 - 5 = \(x^2-\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c. x2 - 7 = \(x^2-\left(\sqrt{7}\right)^2=\left(x+\sqrt{7}\right)\left(x-\sqrt{7}\right)\)
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\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\\ \Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
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Tại mk lười dùng delta nên bn làm delta cũng tương tự vậy nha!
Ta có: x2 - 4x + 5m - 2 = 0
\(\Leftrightarrow\) x2 - 4x + 4 + 5m - 6 = 0
\(\Leftrightarrow\) (x - 2)2 = 6 - 5m
\(\Leftrightarrow\) x - 2 = \(\pm\)\(\sqrt{6-5m}\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x_1=\sqrt{6-5m}+2\\x_2=-\sqrt{6-5m}+2\end{matrix}\right.\)
Ta có: x12 . x2 + x1 . x22 = 12
\(\Leftrightarrow\) (\(\sqrt{6-5m}+2\))2. \(\left(-\sqrt{6-5m}+2\right)\) + \(\left(\sqrt{6-5m}+2\right)\) \(\left(-\sqrt{6-5m}+2\right)^2\) = 12
\(\Leftrightarrow\) (4 - 6 + 5m)(\(\sqrt{6-5m}+2-\sqrt{6-5m}+2\)) = 12
\(\Leftrightarrow\) (-2 + 5m).4 = 12
\(\Leftrightarrow\) -2 + 5m = 3
\(\Leftrightarrow\) m = 1
Vậy ...
Chúc bn học tốt!
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Ta có: \(2x^2+x=2\)
\(\Leftrightarrow2x^2+x-2=0\)
\(Δ=1^2-4\cdot2\cdot\left(-2\right)=1+16=17\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{17}}{4}\\x_2=\dfrac{-1+\sqrt{17}}{4}\end{matrix}\right.\)
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a: \(3\sqrt{200}=3\cdot10\sqrt{2}=30\sqrt{2}\)
b: \(-5\sqrt{50a^2b^2}=-5\cdot5\sqrt{2a^2b^2}\)
\(=-25\cdot\left|ab\right|\cdot\sqrt{5}\)
c: \(-\sqrt{75a^2b^3}\)
\(=-\sqrt{25a^2b^2\cdot3b}=-5\left|ab\right|\cdot\sqrt{3b}\)
\(x^2.2x=x.x.2.x=x.x.x.2=x^3.2\)
😋