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Lên lớp 10 cái đó được tính vào phần toán đại đấy

a: \(\left\{{}\begin{matrix}x+3y=11\\3x-y=9-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+3y=11\\3x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\9x+3y=27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+3y-x-3y=27-11\\x+3y=11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}8x=16\\3y=11-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{11-x}{3}=\dfrac{11-2}{3}=\dfrac{9}{3}=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+10y-3x=-1\\2x+4-3x+15y=-12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+10y=-1\\-x+15y=-16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-2x+30y=-32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+10y-2x+30y=-1+\left(-32\right)\\x-15y=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}40y=-33\\x=15y+16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=15\cdot\dfrac{-33}{40}+16=\dfrac{29}{8}\end{matrix}\right.\)

a) 

\(\left\{{}\begin{matrix}x+3y=11\\3x-y=9-2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\3x+y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+9y=33\\3x+y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}8y=24\\3x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\3x+3=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{6}{3}=2\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-15y-12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\x-15y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\2x-30y=32\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}40y=-33\\x-15y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x+\dfrac{99}{8}=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=16-\dfrac{99}{8}=\dfrac{29}{8}\end{matrix}\right.\)

\(\dfrac{x+100}{4}+\dfrac{x+99}{5}=\dfrac{x+98}{6}+\dfrac{x+97}{7}\)

=>\(\left(\dfrac{x+100}{4}+1\right)+\left(\dfrac{x+99}{5}+1\right)=\left(\dfrac{x+98}{6}+1\right)+\left(\dfrac{x+97}{7}+1\right)\)

=>\(\dfrac{x+104}{4}+\dfrac{x+104}{5}=\dfrac{x+104}{6}+\dfrac{x+104}{7}\)

=>\(\left(x+104\right)\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)=0\)

=>x+104=0

=>x=-104

\(\dfrac{x+100}{4}+\dfrac{x+99}{5}=\dfrac{x+98}{6}+\dfrac{x+97}{7}\\ \dfrac{x+100}{4}+\dfrac{x+99}{5}-\dfrac{x+98}{6}-\dfrac{x+97}{7}=0\\ \left(\dfrac{x+100}{4}+1\right)+\left(\dfrac{x+99}{5}+1\right)-\left(\dfrac{x+98}{6}+1\right)-\left(\dfrac{x+97}{7}+1\right)=0\\ \dfrac{x+104}{4}+\dfrac{x+104}{5}-\dfrac{x+104}{6}-\dfrac{x+104}{7}=0\\ \left(x+104\right)\cdot\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)=0\)

Vì \(\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)\ne0\) nên:

\(x+104=0\\ x=-104\)

Vậy \(x=-104\)

Thay x=-2 và y=3 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}2a\cdot\left(-2\right)+3b\cdot3=5\\-3a\cdot\left(-2\right)+2b\cdot3=30\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-4a+9b=5\\6a+6b=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-12a+27b=15\\12a+12b=60\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-12a+27b+12a+12b=15+60\\a+b=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}39b=75\\a=5-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{75}{39}=\dfrac{25}{13}\\a=5-\dfrac{25}{13}=\dfrac{40}{13}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+2y=1+3\\2x-3y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=4\\7y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=4\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=4-2=2\\y=1\end{matrix}\right.\)

Vậy: ...

\(\left\{{}\begin{matrix}x+2y=1+3\\2x-3y=1\end{matrix}\right.\)⇒ \(\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}2x+4y=8\\2x+4y-\left(2x-3y\right)=8-1\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}2x+4y=8\\2x+4y-2x+3y=7\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}2x+4y=8\\\left(2x-2x\right)+\left(4y+3y\right)=7\end{matrix}\right.\)

⇒  \(\left\{{}\begin{matrix}2x+4y=8\\0+7y=7\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}2x+4y=8\\y=1\end{matrix}\right.\)

Thay y = 1 vào biểu thức 2\(x\) + 4y = 8 ta có: 2\(x\) + 4.1 = 8 

⇒ 2\(x\) + 4 = 8 ⇒ 2\(x\) = 4 ⇒ \(x\) = 4: 2 ⇒ \(x\) = 2

Vậy \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a. \(2x\left(3x+1\right)-7\left(3x-1\right)=0\)

\(\Leftrightarrow6x^2+2x-21x+7=0\)

\(\Leftrightarrow6x^2-19x+7=0\)

\(\Delta=19^2-4.6.7=193>0\)

\(\Rightarrow\) Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{19+\sqrt{193}}{12}\\x_2=\dfrac{19-\sqrt{193}}{12}\end{matrix}\right.\)

b. \(4x^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow-\left(4x+1\right)=0\)

\(\Leftrightarrow4x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{4}\)

c. \(4x^2-4x+1=0\)

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

\(\Leftrightarrow2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

d. \(x\left(x+2\right)-7x-14=0\)

\(\Leftrightarrow x\left(x+2\right)-7\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)

#$\mathtt{Toru}$

Nửa chu vi của hình chữ nhật là: 60:2=30(m)

Chiều rộng của hình chữ nhật là: (30-6):2=12(m)

Chiều dài của hình chữ nhật là: 30 - 12 = 18(m)

Diện tích mảnh đất là: 18 x 12 = 216 (m2)

Đáp số: 216m2

                      Giải:

Nửa chu vi của mảnh vườn hình chữ nhật là:

                  60 : 2 = 30 (m)

Gọi chiều rộng lúc đầu của mảnh vườn hình chữ nhật là \(x\) (m); 30 > \(x\) > 0

Chiều dài lúc đầu của mảnh vườn hình chữ nhật là: 30 - \(x\) (m)

Chiều dài của mảnh vườn hình chữ nhật lúc sau là:

  30 - \(x\) + 2 =  (30 + 2) - \(x\) = 32 - \(x\) (m)

Chiều rộng của hình chữ nhật lúc sau là: \(x\) + 6 (m)

Diện tích của hình chữ nhật lúc sau là:

                 (32 - \(x\))(\(x\) + 6)   (m²)  

Diện tích của mảnh  vườn hình chữ nhật lúc đầu là:  (30 - \(x\)) x \(x\) = 30\(x\) - \(x^2\) (m²)

Theo bài ra ta có phương trình: 

        (32 - \(x\))(\(x\) + 6) - (30\(x\) - \(x^2\)) = 96

          32\(x\) + 192 - \(x^2\) - 6\(x\) - 30\(x\) + \(x^2\) = 96

          (32\(x\) - 6\(x\) - 30\(x\)) + 192 - (\(x^2\) - \(x^2\)) = 96

            (26\(x\) - 30\(x\)) + 192 + 0 = 96

                    - 4\(x\) + 192 = 96

                                 4\(x\) = 192 - 96

                                 4\(x\) = 96

                                \(x\) = 96 : 4

                                \(x\) = 24

Chiều dài ban đầu của hình chữ nhật là: 30 - 24 = 6 (m)

     6 < 24

Chiều dài nhỏ hơn chiều rộng, không có hình chữ nhật nào có kích thước thoả mãn đề bài. 

3)

a) Ta có:

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}\\ =\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2a+2b+2c}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}\\ =\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)

\(=>\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) 

b) 

\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2024^2}+\dfrac{1}{2025^2}}\left(1\right)\\ =\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{\left(-3\right)^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{\left(-4\right)^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2024^2}+\dfrac{1}{\left(-2025\right)^2}}\)

Theo câu a \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) khi \(a+b+c=0\)

Mà: \(\left\{{}\begin{matrix}1+2+\left(-3\right)=0\\1+3+\left(-4\right)=0\\....\\1+2024+\left(-2025\right)=0\end{matrix}\right.\)  

=> \(\left(1\right)=\left|1+\dfrac{1}{2}+\dfrac{1}{-3}\right|+\left|1+\dfrac{1}{3}+\dfrac{1}{-4}\right|+...+\left|1+\dfrac{1}{2024}+\dfrac{1}{-2025}\right|\)  

Mà: \(\left\{{}\begin{matrix}1+\dfrac{1}{2}\cdot\dfrac{1}{-3}>0\\1+\dfrac{1}{3}+\dfrac{1}{-4}>0\\...\\1+\dfrac{1}{2024}+\dfrac{1}{-2025}>0\end{matrix}\right.\) 

=> \(\left(1\right)=1+\dfrac{1}{2}+\dfrac{1}{-3}+1+\dfrac{1}{3}+\dfrac{1}{-4}+...+1+\dfrac{1}{2024}+\dfrac{1}{-2025}\\ =\left(1+1+...+1\right)+\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2024}-\dfrac{1}{2024}\right)-\dfrac{1}{2025}\\ =2023+\dfrac{1}{2}-\dfrac{1}{2025}\)

4)

\(\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{14-8\sqrt{3}}}-3\sqrt{2\left(1-\sqrt{-2\sqrt{3}+4}\right)+2\sqrt{4+2\sqrt{3}}}\\ =\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{4-2\sqrt{3}}\right)+2\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}\\ =\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{\left(2\sqrt{2}-\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}\right)+2\sqrt{\left(\sqrt{3}+1\right)^2}}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}-1\right)^2}\right)+2\left(\sqrt{3}+1\right)}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2\left(1-\sqrt{3}+1\right)+2\sqrt{3}+2}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2-2\sqrt{3}+2+2\sqrt{3}+2}\\ =3\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{6}\right)\cdot\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{6}\\ =3\sqrt{2}\cdot\sqrt{3}-3\sqrt{6}\\ =3\sqrt{6}-3\sqrt{6}\\ =0\)

a: Xét ΔABC có AD là phân giác

nên \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)

=>\(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{120}{2}\right)=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos60\)

=>\(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot\dfrac{1}{2}=\dfrac{AB\cdot AC}{AB+AC}\)

=>\(\dfrac{1}{AD}=\dfrac{AB+AC}{AB\cdot AC}=\dfrac{1}{AB}+\dfrac{1}{AC}\)

b:  \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)

=>\(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos45=\dfrac{2\cdot AB\cdot AC\sqrt{2}}{2\left(AB+AC\right)}=\dfrac{AB\cdot AC\cdot\sqrt{2}}{AB+AC}\)

=>\(\dfrac{1}{AD}=\dfrac{AB+AC}{AB\cdot AC}\cdot\dfrac{1}{\sqrt{2}}\)

=>\(\dfrac{\sqrt{2}}{AD}=\dfrac{AB+AC}{AB\cdot AC}=\dfrac{1}{AB}+\dfrac{1}{AC}\)

c: \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)

=>\(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{60}{2}\right)\)

=>\(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos30=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot\dfrac{\sqrt{3}}{2}\)

=>\(\dfrac{AD}{\sqrt{3}}=\dfrac{AB\cdot AC}{AB+AC}\)

=>\(\dfrac{\sqrt{3}}{AD}=\dfrac{AB+AC}{AB\cdot AC}=\dfrac{1}{AB}+\dfrac{1}{AC}\)