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\(A=x^3-5x^2+3x+9\\ =x^3+x^2-6x^2-6x+9x+9\\ =x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\\ =\left(x^2-6x+9\right)\left(x+1\right)\\ =\left(x-3\right)^2\left(x+1\right)\)
A = \(x^3\) - 5\(x^2\) + 3\(x\) + 9
A = \(x^3\) - 3\(x^2\) - \(x^2\) - \(x^2\) + 3\(x\) + 9
A = (\(x^3\) - 3\(x^2\)) - (\(x^2\) - 3\(x\)) - (\(x^2\) - 9)
A = \(x^2\)(\(x\) - 3) - \(x\)(\(x\) - 3) - (\(x\) - 3)(\(x\) + 3)
A = (\(x\) - 3)(\(x^2\) - \(x\) - \(x\) - 3)
A = (\(x\) - 3)[\(x^2\) - (\(x+x\)) - 3]
A = (\(x\) - 3)[\(x^2\) - 2\(x\) - 3]
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{2a+4}{a\sqrt{a}-1}+\dfrac{\sqrt{a}+2}{a+\sqrt{a}+1}-\dfrac{2}{\sqrt{a}-1}\left(a\ne1;a\ge0\right)\\ =\dfrac{2a+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{2a+4+\left(a+2\sqrt{a}-\sqrt{a}-2\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\)
\(a=3-2\sqrt{2}=\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2=\left(\sqrt{2}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\left(3-2\sqrt{2}\right)+\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}+\sqrt{2}-1+1}=\dfrac{\sqrt{2}-1}{3-\sqrt{2}}=\dfrac{2\sqrt{2}-1}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4+7+10+...+340\)
Số số hạng của dãy \(4+7+10+...+340\) là:
\(\left(340-4\right):3+1=113\) (số)
Giá trị tổng \(4+7+10+...+340\) là:
\(4+7+10+...+340=\left(340+4\right)\times113:2=19436\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x-\dfrac{3}{5}=\dfrac{-7}{10}\\ 3x=\dfrac{-7}{10}+\dfrac{3}{5}\\ 3x=\dfrac{-7}{10}+\dfrac{6}{10}\\ 3x=\dfrac{-1}{10}\\ x=\dfrac{-1}{30}\)
___________________
\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\\ \dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\\ \dfrac{1}{3}:x=\dfrac{9}{15}-\dfrac{10}{15}\\ \dfrac{1}{3}:x=\dfrac{-1}{15}\\ x=\dfrac{1}{3}:\dfrac{-1}{15}\\ x=-5\)
\(3x-\dfrac{3}{5}=-\dfrac{7}{10}\)
\(3x\) \(=-\dfrac{7}{10}+\dfrac{3}{5}\)
\(3x\) \(=-\dfrac{1}{10}\)
\(x\) \(=-\dfrac{1}{10}:3\)
\(x\) \(=-\dfrac{1}{30}\)
Vậy \(x=-\dfrac{1}{30}\)
\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\)
\(\dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{15}\)
\(x=\dfrac{1}{3}:-\dfrac{1}{15}\)
\(x=-5\)
Vậy \(x=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2-3x+1\\ =2x^2-x-2x+1\\ =\left(2x^2-x\right)-\left(2x-1\right)\\ =x\left(2x-1\right)-\left(2x-1\right)\\ =\left(2x-1\right)\left(x-1\right)\)
\(2x^2-3x+1\\ =2x^2-2x-x+1\\ =2x\left(x-1\right)-\left(x-1\right)\\ =\left(2x-1\right)\left(x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}-\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{\sqrt{1^2+2\cdot1\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\\ =\dfrac{2}{\sqrt{2}}\\ =\sqrt{2}\)
b)
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ =\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot1+1^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\\ =\dfrac{0}{\sqrt{2}}\\ =0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a)8^2\cdot8^3\\ =8^{2+3}\\ =8^5\\ b)12^5:12^2\\ =12^{5-2}\\ =12^3\\ c)7^4:7\\ =7^{4-1}\\ =7^3\\ d)9^{15}\cdot9\\ =9^{15+1}\\ =9^{16}\\ e)64:4^2\\ =4^3:4^2\\=4^{3-2}\\ =4\\ f)216\cdot6^{20}\\ =6^3\cdot6^{20}\\ =6^{3+20}\\ =6^{23}\\ g)64:16\\ =2^6:2^4\\ =2^{6-4}\\ =2^2\\ h)a^2\cdot a^7:a=a^{2+7-1}\\ =a^8\)
Bài 2:
a: \(3^x=81\)
=>\(3^x=3^4\)
=>x=4
b: \(\left(3x-5\right)^2=49\)
=>\(\left[{}\begin{matrix}3x-5=7\\3x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=12\\3x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
c: \(6^{x-5}=36\)
=>\(6^{x-5}=6^2\)
=>x-5=2
=>x=5+2=7
d: \(\left(7-2x\right)^3=27\)
=>7-2x=3
=>2x=7-3=4
=>x=4/2=2
Bài 1:
1; Ta có Oy là tia chung của hai góc xOy và yOz
Mặt khác: \(\widehat{xOy}\) + \(\widehat{yOz}\) = 400 + 200 = 600 = \(\widehat{xOz}\)
Vậy tia Oy nằm giữa hai tia Ox và Oz (đúng)
2; Tam giác PQR có các cạnh lần lượt là: PQ; QR; PR
Vậy tam giác PQR là hình gồm ba đoạn PQ; QR; RP đúng
3; Nếu tia Ox nằm giữa hai tia Oy và Oz thì
\(\widehat{yOz}\) = \(\widehat{yOx}\) + \(\widehat{xOz}\)
Vậy \(\widehat{xOy}\) + \(\widehat{zOy}\) = \(\widehat{xOz}\) (sai)
4; Hai góc kề bù là hai góc có một cạnh chung và hai cạnh còn lại là hai tia đối nhau.
Vậy hai góc kề bù là hai góc có hai cạnh là hai tia đối nhau là sai
Bài 2
1; Số nghich đảo của 0,25 là:
1 : 0,25 = 4
Chọn B.4
2; 60% của 55 là: 55 x 60 : 100 = 33
Chọn A.33