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a, Ta có \(\left(x+y\right)^2=\left(x+y\right)\left(x+y\right)=x^2+2xy+y^2\)
\(\Rightarrow x^2+y^2+2xy=49\)
\(\Rightarrow x^2+y^2=49-2\left(-18\right)\)\(=85\)
b, \(\left(x-y\right)^2=\left(x-y\right)\left(x-y\right)=x^2-2xy+y^2\)\(=\left(x^2+y^2\right)-2\left(-18\right)\)\(=85+36=121\)
\(\Leftrightarrow\left(x-y\right)^2=121\Rightarrow x-y=11\)
Ta có \(\hept{\begin{cases}x-y=11\\x+y=7\end{cases}}\)
Trừ xuống : \(-2y=4\Rightarrow x=-2\)
Mà \(x+y=7\Rightarrow x-2=7\Rightarrow x=9\)
Vậy \(x=9\); \(y=-2\)
bạn Thùy Linh ơi sai đề rồi bạn. Dù sao cũng cảm ơn nha!
a, \(\left[x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x\left(x^2-16\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x^3-16x-x^2-1\right]x^2-1\)
\(=x^5-16x^3-x^4-x^2-1\)
b, \(\left(y-3\right)y+3y^2+9-y^2+2\left(y^2-2\right)\)
\(=y^2-3y+3y^2+9-y^2+2y^2-4\)
\(=5y^2-3y+5\)
c, \(\left(x+y\right)\left(x^2x^2-xy+y^2\right)\)
\(=x^5-x^2y+xy^2+x^4y-xy^2+y^3\)
d, \(\left(\dfrac{1}{2}xy+\dfrac{3}{4}y\right).\dfrac{1}{2}xy-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}x^2y^2+\dfrac{3}{8}xy^2-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}y.\left(x^2y+\dfrac{3}{2}xy-3\right)\)
Chúc bạn học tốt!!!
Tớ làm lần lượt nhé.
Ta có:\(\frac{3}{x-1}=\frac{4}{y-2}=\frac{5}{z-3}\)
\(\Rightarrow\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta được:
\(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}=\frac{\left(x-1\right)+\left(y-2\right)+\left(z-3\right)}{3+4+5}=\frac{\left(x+y+z\right)-\left(1+2+3\right)}{12}=\frac{18-6}{12}=1\)
\(\Rightarrow\frac{x-1}{3}=1\Rightarrow x=4\)
\(\frac{y-2}{4}=1\Rightarrow y=6\)
\(\frac{z-3}{5}=1\Rightarrow z=3\)
\(\frac{x-y}{2}=\frac{x+y}{12}=\frac{xy}{200}=\frac{x-y+x+y}{2+12}=\frac{2x}{14}=\frac{x}{7}=k\)
\(\Rightarrow x=7k\left(1\right);x+y=12k\left(2\right);xy=200k\left(3\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow y=12k-7k=5k\)
\(\Rightarrow xy=5k\cdot7k=35k^2\left(4\right)\)
Từ \(\left(3\right);\left(4\right)\Rightarrow200k=35k^2\Leftrightarrow200=35k\Leftrightarrow k=\frac{200}{35}\)
\(\Rightarrow x=7\cdot\frac{200}{35}=40\)
\(y=5\cdot\frac{200}{35}=\frac{1000}{35}\)
P/S:số khá xấu.sợ sai.nhưng cách làm là như vậy.
a. \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(xy=54\Rightarrow2k3k=54\Rightarrow6k^2=54\Rightarrow k^2=9\Rightarrow k\in\left\{3;-3\right\}\)
\(k=3\Rightarrow x=6;y=9\)
\(k=-3\Rightarrow x=-6;y=-9\)
b.\(\frac{x}{5}=\frac{y}{3}=k\Rightarrow x=5k;y=3k\)
\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\Rightarrow25k^2-9k^2=4\)
\(\Rightarrow16k^2=4\Rightarrow k^2=\frac{1}{4}\Rightarrow k\in\left\{\frac{1}{2};-\frac{1}{2}\right\}\)
\(k=\frac{1}{2}\Rightarrow x=\frac{5}{2};y=\frac{3}{2}\)
\(k=-\frac{1}{2}\Rightarrow x=\frac{-5}{2};y=\frac{-3}{2}\)
c.\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{2}.\frac{1}{5}=\frac{y}{3}.\frac{1}{5}\Rightarrow\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{3}=\frac{z}{7}.\frac{1}{3}\Rightarrow\frac{y}{15}=\frac{z}{21}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
\(\Rightarrow x=20,y=30,z=42\)
d.\(\frac{x^2}{9}=\frac{y^2}{16}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)
\(\Rightarrow x^2=36\Rightarrow x\in\left\{6;-6\right\};y^2=64\Rightarrow y\in\left\{8;-8\right\}\)
a, \(A=x^3-x^2y+3x^2-xy+y^2-4y+x+2\)
\(=x^3-x^2y+3x^2-\left(xy-y^2+3y\right)-y+x+3-1\)
\(=x^2\left(x-y+3\right)-y\left(x-y+3\right)+\left(x-y+3\right)-1\)
Thay x-y+3=0 vào A
\(A=x^2.0-y.0+0-1=-1\)
b, \(B=x^3-2x^2y+3x^2+xy^2-3xy-2y+2x+4\)
\(=x^3-x^2y-x^2y+3x^2+xy^2-3xy-2y+2x+4\)
\(=x^3-x^2y+3x^2-x^2y+xy^2-3xy+2x-2y+6-2\)
\(=x^2\left(x-y+3\right)-xy\left(x-y+3\right)+2\left(x-y+3\right)-2\)
Thay x-y+3=0 vào B
\(B=x^2.0-xy.0+2.0-2=-2\)
1 Ta có x -24 = y
Suy ra x - y = 24
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
x/7 = y/3 = x-y/7-3 =24/4=6
suy ra x= 42
y = 18