rút gọn
a)\(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}\left(a,b\in Q;a\ne1;b\ne-1\right)\)
b)\(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\left(a,b\in Q,a\ne\frac{1}{2};b\ne-1\right)\)
các bạn giúp mình nha. Mình cảm ơn nhiều
Những câu hỏi liên quan
Rút gọn
\(\frac{1}{a\left(a-b\right)\left(a-c\right)}+\frac{1}{b\left(b-c\right)\left(b-a\right)}+\frac{1}{c\left(c-a\right)\left(c-b\right)}\)
Rút gọn biểu thức \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{b-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{c-b+b-a+a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
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Rút gọn biểu thức sau :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
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rút gọn bt biết a,b,c dương ; ab=1 và a+b khác 0
\(\frac{1}{\left(a+b\right)^3}.\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}.\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
Xem thêm câu trả lời
1.Tìm GTNN của \(B=\frac{|x|+2020}{2019}\)
2.Rút gọn
a,\(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}\)(a,b\(\in Q;a\ne1;b\ne-1)\)
b,\(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\)\(\left(a,b\in Q;a\ne\frac{1}{2};b\ne-1\right)\)
rút gọn:
\(\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}+1\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)1
Tớ giải bừa
\(\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}+1\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)
\(=\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}\right)^2-1^2\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=2ab-2b^2\)
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\(\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}+1\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)
\(=\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}^2-1\right)\)
\(=\left(a-b\right)^2\left(\left|\frac{a+b}{a-b}\right|-1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)(Vì \(\frac{a+b}{a-b}\)nằm trong dấu căn ban đầu)
\(=\frac{\left(a-b\right)^2\left(a+b\right)}{a-b}-\left(a-b\right)^2\)
\(=a^2-b^2-a^2+2ab-b^2\)
\(=2ab-2b^2\)
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Cho biểu thức:
P= \(\frac{a^2}{\left(a+b\right)\left(1-b\right)}-\frac{b^2}{\left(a+b\right)\left(1+a\right)}-\frac{a^2b^2}{\left(1+a\right)\left(1-b\right)}\)
a) Rút gọn P
b) Tìm cặp số nguyên (a;b) sao cho P=3
a) Điều kiện : \(a\ne-b;b\ne1;a\ne-1\)
\(P=\frac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2+b^2-a^2b^2+a-b-ab}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2\left(1-b^2\right)-\left(1-b^2\right)+a\left(1-b\right)+\left(1-b\right)}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(1-b\right)\left(a^2+a^2b-1-b+a+1\right)}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2+a^2b+a-b}{1+a}\)
\(P=\frac{a\left(a+1\right)+b\left(a-1\right)\left(a+1\right)}{1+a}\)
\(P=\frac{\left(a+1\right)\left(a+ab-b\right)}{1+a}\)
P = a + ab - b
b)
P = 3
<=> a + ab - b = 3
<=> a(b+1) - (b+1) +1 - 3 = 0
<=> (b+1)(a-1) = 2
Ta có bảng sau với a, b nguyên
| b+1 | 1 | 2 | -1 | -2 |
| a-1 | 2 | 1 | -2 | -1 |
| b | 0 | 1 | -2 | -3 |
| a | 3 | 2 | -1 | 0 |
| so với đk | loại | loại |
Vậy (a;b) \(\in\){ (3; 0) ; (0; -3)}
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Rút gọn biểu thức:
A=\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
Toán violympic nhé trình bày cách làm giúp mik vs
Cho ab + bc + ca = 1
Rút gọn: P =\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}-\frac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)