\(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)
\(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)
Ta có : \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}\ge0\forall x\\\left(y+\frac{1}{3}\right)^{40}\ge0\forall y\end{cases}}\Rightarrow\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}\ge0\forall x;y\)
Khi đó \(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)
<=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y+\frac{1}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)
Vậy x = 1/2 ; y = -1/3
Ta có: \(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}\ge0\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}=0\\\left(y+\frac{1}{3}\right)^{40}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)
Vì \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}\\\left(y+\frac{1}{3}\right)^{40}\end{cases}}\ge0\forall x,y\Rightarrow\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}\ge0\)
Dấu "=" xảy ra <=> x = 1/2 ; y = -1/3
Vậy x = 1/2 ; y = -1/3
Tìm x,y biết a) \(\frac{x}{18}=\frac{y}{15}\) và x-y=-30 b) 7x=9y và 10x-8y=68 c) \(\left(x-\frac{1}{2}\right)^{^{50}}+\left(y+\frac{1}{3}\right)^{^{40}}=0\)
Cho x và y là hai số khác 0 và thỏa mãn x+y khác 0. Chứng minh rằng:
\(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3y^3}\)
Tìm x, y biết:
b) 7x = 9y và 10x - 8y = 68
c) \(\left(x-\frac{1}{2}\right)^{50}\)+ \(\left(y+\frac{1}{3}\right)^{40}\)= 0
b) \(7x=9y\) và \(10x-8y=68\)
Có: \(7x=9y\Leftrightarrow\frac{x}{9}=\frac{y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{9}=\frac{y}{7}=\frac{10x-8y}{90-56}=\frac{68}{34}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.9\\y=2.7\end{cases}}\Rightarrow\hept{\begin{cases}x=18\\y=14\end{cases}}\)
b) Ta có: 7x = 9y => x = 9/7y
Lại có: 10x - 8y = 68
=> 10.9/7.y - 8y = 68
=> 90/7.y - 56/7.y = 68
=> 34/7.y = 68
=> y = 68 : 34/7 = 14
=> x = 9/7.14 = 18
c) Vì (x - 1/2)50 > hoặc = 0; (y + 1/3)40 > hoặc = 0
Mà (x - 1/2)50 + (y + 1/3)40 = 0
=> (x - 1/2)50 = 0; (y + 1/3)40 = 0
=> x - 1/2 = 0; y + 1/3 = 0
=> x = 1/2; y = -1/3
c) \(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)
Có: \(\left(x-\frac{1}{2}\right)^{50}\ge0;\left(y+\frac{1}{3}\right)^{40}\ge0\)
Theo bài ra: \(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}=0\\\left(y+\frac{1}{3}\right)^{40}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+\frac{1}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)
Giải hệ phương trình:\(\left\{{}\begin{matrix}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{2}\left(xy+3x+2y+6\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{2}x+y+3=50\\x+y-2=32\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{2}x+5=18\\x+y-2=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
vậy hệ phương trình có nghiệm(x;y)=(26;8)
Tìm x biết:
a) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
b) \(\left(\frac{1}{2}x-3\right).\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
c) \(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
d) \(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
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a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
Giải phương trình:
a,\(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
b,\(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
c,\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+3}\)
Tìm x biết
a)\(\frac{x+1}{x-4}>0\)
b)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
c)\(\left(x+2\right)\left(x-3\right)< 0\)
d)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|\le0\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
\(\Rightarrow\frac{x-4}{x-4}+\frac{5}{x-4}>0\)
\(\Rightarrow1+\frac{5}{x-4}>0\)
\(\Rightarrow\frac{5}{x-4}>-1\)
\(\Rightarrow\frac{-5}{-x+4}>-\frac{5}{5}\)
\(\Rightarrow-x+4< -5\)
\(\Rightarrow-x< -9\)
\(\Rightarrow x>9\)
giải hệ phương trình:
1) \(\hept{\begin{cases}2\left(x+y\right)+3\left(x+y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{cases}}\)
2)\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12_{ }\end{cases}}\)
3) \(\hept{\begin{cases}\frac{2y-5x}{3}+5=\frac{y+27}{4}-2x\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{cases}}\)
4)\(\hept{\begin{cases}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{cases}}\)
5)\(\hept{\begin{cases}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{cases}}\)
Những bài còn lại chỉ cần phân tích ra rồi rút gọn là được nha. Bạn tự làm nha!
Đặt \(\hept{\begin{cases}x+y=a\\x-y=b\end{cases}}\)\(\Rightarrow\)ta có hệ \(\hept{\begin{cases}2a+3b=4\\a+2b=5\end{cases}}\Rightarrow\hept{\begin{cases}a=-7\\b=6\end{cases}}\)Từ đó ta có \(\hept{\begin{cases}x+y=-7\\x-y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{13}{2}\end{cases}}\)PS: Cái đề chỗ 3(x+y) phải thành 3(x-y) chứ
2) Từ hệ ta có \(\hept{\begin{cases}20x-6y=66\\-3x=-9\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)