Cho x-y=7 Tính
\(H=x^2\cdot\left(x+1\right)-y^2\cdot\left(y-1\right)+xy-3xy\cdot\left(x-y+1\right)-95\)
Rút gọn: \(\frac{x^2}{\left(x+y\right)\cdot\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\cdot\left(1+x\right)}-\frac{x^2\cdot y^2}{\left(x+1\right)\cdot\left(1-y\right)}\)
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
Tính các biểu thức sau :
\(\dfrac{a.\left(7\cdot x^2+11\cdot y^2\right)}{\left(14\cdot x^{12}-11\cdot y^2\right)}.\left(\dfrac{x}{11}\right)=\left(\dfrac{y}{7}\right)\)
1)
a, Cho x,y với xy lớn hơn hoặc bằng 0. Cm \(\left(x^2-y^2\right)^2\) lớn hơn hoặc bằng \(\left(x-y\right)^2\)
b, Cho \(x\cdot y\cdot z=1\) và \(x+y+z>\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\). Cm \(\left(x-1\right)\cdot\left(y-1\right)\cdot\left(z-1\right)>0\)
\(\left(x^2-y^2\right)^2=\left(x-y\right)^2\left(x+y\right)^2\) \(\Rightarrow\left\{{}\begin{matrix}x;y>0\\x+y< 1\end{matrix}\right.\)=> dccm sai = > người ra đề sai họăc người chép đề sai ;
\(\left(x-3\right)\cdot\left(y-2\right)=7\)
\(\left(x-1\right)\cdot\left(y-1\right)=2\)
\(\left(x-1\right)\cdot\left(y-2\right)=2\)
>> Với toán lớp 6 chắc đề bài là tìm x,y nhỉ ? . Lần sau bạn nhớ viết tên đề bài nhé ;) <<
a) \((x−3).(y−2)=7\)
\(\Rightarrow\left(x\text{−}3\right)\inƯ\left(7\right)\)
\(\Rightarrow x\text{−}3\in\left\{1;\text{−}1;7;\text{−}7\right\}\)
Ta có bảng sau :
\(x\text{−}3\) | \(1\) | \(−1\) | \(7\) | \(−7\) |
\(x\) | \(4\) | \(2 \) | \(10\) | \(\text{−}4\) |
\(y−2\) | 7 | −7 | 1 | −1 |
\(y\) | 9 | −5 | 3 | 1 |
Vậy .....
b) \((x−1).(y−1)=2\)
\(\Rightarrow\left(x\text{−}1\right)\inƯ\left(2\right)\)
\(\Rightarrow x\text{−}1\in\left\{1;\text{−}1;2;\text{−}2\right\}\)
Ta có bảng sau :
x−1 | 1 | −1 | 2 | −2 |
x | 2 | 0 | 3 | −1 |
y−1 | 2 | −2 | 1 | −1 |
y | 3 | −1 | 2 | 0 |
Vậy ......
c) \((x−1).(y−2) = 2\)
\(\Rightarrow x\text{−}1\inƯ\left(2\right)\)
\(\Rightarrow x\text{−}1\in\left\{1;\text{−}1;2;\text{−}2\right\}\)
Ta có bảng sau :
x−1 | 1 | −1 | 2 | −2 |
x | 2 | 0 | 3 | −1 |
y−2 | 2 | −2 | 1 | −1 |
y | 4 | 0 | 3 | 1 |
Vậy ...
chứng minh \(x^2\cdot\left(1+y^2\right)+y^2\cdot\left(1+z^2\right)+z^2\cdot\left(1+x^2\right)\ge6\cdot x\cdot y\cdot z\)
\(a=x\cdot y+\sqrt{\left(1+x^2\right)\cdot\left(1+y^2\right)}\) \(b=x\cdot\sqrt{1+y^2}+y\cdot\sqrt{1+x^2}\) với xy>0 tính b theo a
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow a^2-b^2=1\)
\(\Rightarrow a^2=1+b^2\)
1, Giải hệ phương trình:
\(\hept{\begin{cases}x\cdot\left|x\right|-\left(x+10\right)\cdot\left|x+10\right|=y\cdot\left|y\right|\\y\cdot\left|y\right|-\left(y+10\right)\cdot\left|y+10\right|=z\cdot\left|z\right|\\z\cdot\left|z\right|-\left(z+10\right)\cdot\left|z+10\right|=x\cdot\left|x\right|\end{cases}}\)
Giải hộ mk nhoa mk tick cho !!!!!!!!!
CM các đẳng thức sau:
\(\left[\frac{x+2}{x+1}-\frac{4\cdot\left(y+1\right)}{y+2}\right]:\left[\frac{x^2\cdot\left(y+1\right)}{y+1}-\frac{y^2\cdot\left(x+2\right)}{y+2}\right]=\frac{1}{y-x}\)
cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)