1) G/s 2 điểm đó là \(A\left(-1;y_1\right)\) và \(B\left(2;y_2\right)\)

\(\Rightarrow\hept{\begin{cases}y_1=-\left(-1\right)^2=-1\\y_2=-2^2=-4\end{cases}}\)

\(\Rightarrow A\left(-1;-1\right)\) và \(B\left(2;-4\right)\)

PT đường thẳng đó công thức là \(y=ax+b\Rightarrow\hept{\begin{cases}-a+b=-1\\2a+b=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-1\\b=-2\end{cases}}\)

Vậy PT đường thẳng đó là \(y=-x-2\)

2) 

a) Với m = -1 : \(x^2-2\cdot\left(-1-1\right)x--1-3=0\)

\(\Leftrightarrow x^2+4x-2=0\)

\(\Leftrightarrow\left(x+2\right)^2=6\Rightarrow x=-2\pm\sqrt{6}\)

b) \(\Delta^'=\left[-\left(m-1\right)\right]^2-1\cdot\left(-m-3\right)\)

\(=m^2-2m+1+m+3=m^2-m+4>0\left(\forall m\right)\)

=> PT luôn có 2 nghiệm phân biệt với mọi m

Theo hệ thức viet: \(\hept{\begin{cases}x_1+x_2=2m-2\\x_1x_2=-m-3\end{cases}}\)

Ta có: \(x_1^2+x_2^2=14\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=14\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)=14\)

\(\Leftrightarrow4m^2-8m+4+2m+6-14=0\)

\(\Leftrightarrow4m^2-6m-4=0\)

\(\Leftrightarrow2m^2-3m-2=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}m=2\\m=-\frac{1}{2}\end{cases}}\left(tm\right)\)

Vậy \(m\in\left\{2;-\frac{1}{2}\right\}\)

Gọi vận tốc dòng nước là \(x\left(km/h\right),0< x< 20\).

Vận tốc của cano khi đi xuôi dòng là: \(20+x\left(km/h\right)\).

Vận tốc của cano khi đi ngược dòng là: \(20-x\left(km/h\right)\)

Theo bài ra, ta có phương trình: 

\(\frac{30}{20+x}+\frac{24}{20-x}=3\)

\(\Leftrightarrow\frac{30\left(20-x\right)+24\left(20+x\right)}{\left(20+x\right)\left(20-x\right)}=\frac{3\left(20+x\right)\left(20-x\right)}{\left(20+x\right)\left(20-x\right)}\)

\(\Rightarrow200-10x+160+8x=400-x^2\)

\(\Leftrightarrow x^2-2x-40=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{41}\left(tm\right)\\x=1-\sqrt{41}\left(l\right)\end{cases}}\)

Xét bài toán phụ sau:

Nếu \(a+b+c=0\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)  \(\left(a,b,c\ne0\right)\)

Thật vậy

Ta có: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{a+b+c}{abc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{0}{abc}}\)

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

Bài toán được chứng minh

Quay trở lại, ta sẽ áp dụng bài toán phụ vào bài chính:

Ta có: \(P=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}+...+\sqrt{\frac{1}{2^2}+\frac{1}{779^2}+\frac{1}{801^2}}\)

Vì \(2+1+\left(-3\right)=0\) nên:

\(\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{\left(-3\right)^2}}=\sqrt{\left(\frac{1}{2}+\frac{1}{1}-\frac{1}{3}\right)^2}=\frac{1}{2}+1-\frac{1}{3}\)

Tương tự ta tính được:

\(\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\) ; ... ; \(\sqrt{\frac{1}{2^2}+\frac{1}{799^2}+\frac{1}{801^2}}=\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)

\(\Rightarrow P=\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)

\(=\frac{1}{2}\cdot400+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{799}-\frac{1}{801}\right)\)

\(=200+\frac{800}{801}=\frac{161000}{801}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=161000\\b=801\end{cases}}\)

\(\Rightarrow Q=161000-801\cdot200=800\)

\(a,\sqrt{22-8\sqrt{6}}\)

\(\sqrt{4^2-8\sqrt{6}+\sqrt{6}^2}\)

\(\sqrt{\left(4-\sqrt{6}\right)^2}=\left|4-\sqrt{6}\right|\)

\(4>\sqrt{6}< =>\left|4-\sqrt{6}\right|=4-\sqrt{6}\)

\(b,\sqrt{16-6\sqrt{7}}\)

\(\sqrt{3^2-6\sqrt{7}+\sqrt{7}^2}\)

\(\sqrt{\left(3-\sqrt{7}\right)^2}\)

\(\left|3-\sqrt{7}\right|\)

\(=3-\sqrt{7}\)

\(c,\sqrt{9-4\sqrt{2}}\)

\(\sqrt{9-2.2\sqrt{2}}\)

\(\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\)

\(\sqrt{\left(2\sqrt{2}-1\right)^2}\)

\(2\sqrt{2}>1\)

\(\left|2\sqrt{2}-1\right|=2\sqrt{2}-1\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\frac{5}{4}}=\frac{4}{5}\)

\(tanx=\frac{sinx}{cosx}=\frac{5}{4}\Rightarrow\frac{sin^2x}{cos^2x}=\frac{25}{16}\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{25}{16}\)

\(\Rightarrow16\left(1-cos^2x\right)=25cos^2x\)

\(\Leftrightarrow cos^2x=\frac{16}{41}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{4}{\sqrt{41}}\Rightarrow sinx=\frac{5}{\sqrt{41}}\\cosx=\frac{-4}{\sqrt{41}}\Rightarrow sinx=\frac{-5}{\sqrt{41}}\end{cases}}\)

Đặt \(x=a,1+y=b\).

Ta có: 

\(a^3+b^3=2ab\)

\(\Leftrightarrow a^4+ab^3=2a^2b\)

\(\Leftrightarrow\left(a^2-b\right)^2-b^2=-ab^3\)

\(\Leftrightarrow\left(a^2-b\right)^2=b^2\left(1-ab\right)\)

\(\Leftrightarrow1-ab=\left(\frac{a^2-b}{b}\right)^2\)

Ta có đpcm. 

lớp 9 à

ghê dậy bn pay nick lunnnn

a, \(A=\frac{2a-a^2}{a+3}\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)ĐK : \(a\ne\pm2;-3\)

\(=\frac{2a-a^2}{a+3}\left(\frac{a^2-4a+4-a^2-4a-4-4a^2}{\left(a+2\right)\left(a-2\right)}\right)\)

\(=\frac{a\left(2-a\right)}{a+3}\left(\frac{4a^2+8a}{\left(a+2\right)\left(2-a\right)}\right)=\frac{4a^2\left(a+2\right)}{\left(a+3\right)\left(a+2\right)}=\frac{4a^2}{a+3}\)

b, \(A=1\Rightarrow\frac{4a^2}{a+3}=1\Rightarrow4a^2=a+3\Leftrightarrow4a^2-a-3=0\)

\(\Leftrightarrow\left(a-1\right)\left(4a+3\right)=0\Leftrightarrow a=-\frac{3}{4};a=1\)( tmđk )

Vậy với a = -3/4 ; a = 1 thì A = 1 

c, Để A nhận giá trị dương khi A > 0 \(\Rightarrow\frac{4a^2}{a+3}>0\Rightarrow a+3>0\Leftrightarrow a>-3\)

Để A nhận giá trị âm khi A < 0 \(\Rightarrow\frac{4a^2}{a+3}< 0\Rightarrow a+3< 0\Leftrightarrow a< -3\)