Rút gọn biểu thức:
a, 3(x - 1) - 2/ x + 3 /
b, 2/ x - 3 / - / 4x - 1/
Các bn lm ơn giúp tớ đi
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a) Mình không hiểu đề cho lắm
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2+3\left(4x+1\right)\left(4x-1\right)\)
\(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3\)
\(=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4\)
\(=-77\)
a. \(4x\left(3x-2\right)-3x\left(4x+1\right)\)
\(=12x^2-8x-12x^2-3x\)
\(=-11x\) \(\left(1\right)\)
Thay \(x=-2\) vào \(\left(1\right)\) ta được :
\(-11.\left(-2\right)=22\)
b. \(\left(x+3\right)\left(x-3\right)-\left(x-1\right)^2\)
\(=\left(x^2-9\right)-\left(x^2-2x+1\right)\)
\(=x^2-9-x^2+2x-1\)
\(=2x-10\) \(\left(2\right)\)
Thay \(x=6\) vào \(\left(2\right)\) ta được :
\(2.6-10=2\)
\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-x-12}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x+2}{x+3}\)
\(=\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{x+2}\)
\(=\frac{3x+6}{x-3}\times\frac{1}{x+2}=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\frac{3}{x-3}\)
\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\left(\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\frac{6+3x}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
Bài 1:
\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)
Bài 2:
\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)
B1:
a,\(\left(3x-2\right)\left(x-3\right)=3x^2-9x-2x+6=3x^2-11x+6\)
b,\(\left(2x+1\right)\left(x+3\right)=2x^2+6x+x+3=2x^2+7x+3\)
c,\(\left(x-3\right)\left(3x-1\right)=3x^2-x-9x+3=3x^2-10x+3\)
B2:
1)\(x^2-\left(x+4\right)\left(x-1\right)=x^2-\left(x^2-x+4x-4\right)=x^2-x^2+x-4x+4=-3x+4\)
2)\(x\left(x+2\right)-\left(x-2\right)\left(x+4\right)=x^2+2x-\left(x^2+4x-2x-8\right)\)
\(=x^2+2x-x^2-4x+2x+8=8\)
a: Ta có: \(P=\left(x-1\right)^2-4x\left(x+1\right)\left(x-1\right)+3\)
\(=x^2-2x+1-4x\left(x^2-1\right)+3\)
\(=x^2-2x+4-4x^3+4x\)
\(=-4x^3+x^2+2x+4\)
b: Thay x=-2 vào P, ta được:
\(P=-4\cdot\left(-8\right)+4-4+4=36\)
/ x - 1 / là giá trị tuyệt đối à ?
ko ạ chỉ có x+3 ở câu a thôi