Câu 1:
a, Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=.....=\dfrac{a_9-9}{1}=\dfrac{a_1-1+a_2-2+......+a_9-9}{9+8+7+....+1}\)
\(=\dfrac{\left(a_1+a_2+.....+a_9\right)-\left(1+2+3+....+9\right)}{45}\)
\(=\dfrac{90-45}{45}=1\) (do \(a_1+a_2+a_3+.....+a_9=90\))
\(\Rightarrow\dfrac{a_1-1}{9}=1;\dfrac{a_2-2}{8}=1;.......;\dfrac{a_9-9}{1}=1\)
\(\Rightarrow a_1=a_2=a_3=.....=a_9=10\)
Vậy..............
Chúc bạn học tốt!!!
Câu 2:
b, \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
\(\Rightarrow\left|x.\left(x+2\right)\right|+\left|y^2-9\right|=0\)
Với mọi giá trị của \(x;y\in R\) ta có:
\(\left|x.\left(x+2\right)\right|\ge0;\left|y^2-9\right|\ge0\)
\(\Rightarrow\left|x.\left(x+2\right)\right|+\left|y^2-9\right|\ge0\)
Để \(\left|x.\left(x+2\right)\right|+\left|y^2-9\right|=0\) thì
\(\left\{{}\begin{matrix}\left|x.\left(x+2\right)\right|=0\\\left|y^2-9\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\\y^2=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\y=\pm3\end{matrix}\right.\)
Vậy các cặp (x;y) thoả mãn đề bài là:
\(\left(x;y\right)\in\left\{\left(0;-3\right);\left(0;3\right);\left(-2;-3\right);\left(-2;3\right)\right\}\)
Chúc bạn học tốt!!!
Ý tính tổng c1 mik gọi la câu a , câu b sẽ là ý còn lại nha
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Câu 1 :
a , Ta có : \(\left(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{7}}\right)+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)
= \(\left(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}\right)+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
= \(\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
b , Theo bài ra ta có : \(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-9}{1}\)= \(\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+..+9\right)}{\left(1+2+...+9\right)}=\dfrac{90-\dfrac{9\left(9+1\right)}{2}}{\dfrac{9\left(9+1\right)}{2}}=\dfrac{90-45}{45}=\dfrac{45}{45}=1\)
=> \(\dfrac{a_1-1}{9}=1\) => a1 = 10
tương tự , ta suy ra : a2 =a3=a4 =a5 = a6=a7= a8=a9= 10
Vậy a1=a2 =a3=a4 =a5 = a6=a7= a8=a9= 10
