thu gon bieu thuc
cho A = 1+3+ 32 +33+ ... + 32015
Những câu hỏi liên quan
thu gon bieu thuc
(x+1)^3-x(x^2+3)
thu gon bieu thuc A=(1-1/1+2)*(1-1/1+2+3)...(1-1/1+2+3+...+2016)
cho bieu thuc A = /x-1/ + 3x-7
a. rut gon bieu thuc
b. tinh A khi x=3;x=-5
thu gon bieu thuc sau (2x-1)^3-(3x^2-1)(x-2)-(x+3)^3
Ta có: \(\left(2x-1\right)^3-\left(3x^2-1\right)\left(x-2\right)+\left(x+3\right)^3\)
\(=8x^3-12x^2+6x-1-\left(3x^3-6x^2-x+2\right)+x^3+9x^2+27x+27\)
\(=9x^3-3x^2+33x+26-3x^3+6x^2+x-2\)
\(=6x^3+3x^2+34x+24\)
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cho bieu thuc a^3+2a^2-1/a^3+2a^2+2a+1
a) Rut gon bieu thuc
b)Chung minh rang neu a la so nguyen thi gia tri tim duoc cua bieu thuc o cau a) la 1 phan so toi gian
Cho bieu thuc A=a3+2a3-1/a3+2a3+2a+1
a ,Rut gon bieu thuc
b,CMR neu a la so nguyen thi gia tri cua bieu thuc tim duoc o cau a la mot phan so toi gian.
a) \(A=\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}=\frac{a^2.\left(a+1\right)+\left(a+1\right).\left(a+1\right)}{a^2.\left(a+1\right)+a.\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right).\left(a^2+a-1\right)}{\left(a+1\right).\left(a^2+a+1\right)}=\frac{ }{ }\)\(\frac{a^2+a-1}{a^2+a-1}\)
duyệt đi
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Thu gon bieu thuc:
E=|x-3|+|x2+1|-x (x-1)
thu gon bieu thuc |2x+3|-x=1
giúp mk gấp nha
thanks
\(\left|2x+3\right|-x=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=1-3\\2x+x=-1-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)
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Cho bieu thuc: ( x-1/ x+1 - x-1/x+1) : 2x / 3x - 3
a, Tim dieu kien xac dinh cua bieu thuc P
b, Rut gon bieu thuc P
c, Tim x thuoc z de P nhan gia tri nguyen.
Đề bài sai rồi bạn ! Mình sửa :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
b) \(P=\left(\frac{x-1}{x+1}-\frac{x+1}{x-1}\right):\frac{2x}{3x-3}\)
\(\Leftrightarrow P=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{x^2-2x+1-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-6}{x+1}\)
c) Để P nhận giá trị nguyên
\(\Leftrightarrow\frac{-6}{x+1}\inℤ\)
\(\Leftrightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)
Ta loại các giá trị ktm
\(\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)