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\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{2x+1}{-\left(2x+1\right)}-\dfrac{2x-1}{-\left(2x-1\right)}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{2x}{5x-5}\)
a) \(\left(3x+2\right).\left(x-3\right)-3x.\left(x+\frac{1}{3}\right)\)
\(=3x^2-9x+2x-6-\left(3x^2+x\right)\)
\(=3x^2-9x+2x-6-3x^2-x\)
\(=\left(3x^2-3x^2\right)+\left(-9x+2x-x\right)-6\)
\(=-8x-6.\)
Chúc bạn học tốt!
\(B=\left(3x-2\right)^2-\left(x+2\right).\left(x-2\right)\)
\(=\left(3x-2\right)^2-\left(x^2-2^2\right)\)
\(=9x^2-12x+4-x^2+4\)
\(=8x-12x+8\)
\(C=\left(x+4\right)^2-7x.\left(x-2\right)\)
\(=x^2+8x+16-\left(7x^2-14x\right)\)
\(=x^2+8x+16-7x^2+14x\)
\(=-6x^2+22x+16\)
\(D=-4x.\left(2x-7\right)+\left(x+5\right)^2\)
\(=-8x^2+28x+x^2+10x+25\)
\(=-7x^2+38x+25\)
2 câu dễ làm trước, 2 câu còn lại tối đi học về mới làm được..(giờ bận rồi)
a) ĐẶt \(x^2+3x+1=a\)
\(A=a\left(a-4\right)-5=a^2-4a-5=\left(a-5\right)\left(a+1\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
c)\(C=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt ẩn phụ: \(t=x^2+8x+7\) rồi làm tiếp đi..
Để anh làm nốt vậy.
\(B=\left(x^2+2x\right)^2-2x^2-4x-3\)
\(B=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1-4\)
\(B=\left(x^2+2x-1\right)^2-2^2\)
\(B=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
\(B=\left(x+3\right)\left(x-1\right)\left(x+1\right)^2\)
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\(D=x^2-2xy+y^2-7x+7y+12\)
\(D=\left(x-y\right)^2-7\left(x-y\right)+12\)
\(D=\left(x-y\right)^2-3\left(x-y\right)-4\left(x-y\right)+12\)
\(D=\left(x-y\right)\left(x-y-3\right)-4\left(x-y-3\right)\)
\(D=\left(x-y-3\right)\left(x-y-4\right)\)
Nhầm 1 chút nhé, Bài 1 câu a) ( x^2 - 5x ) . ( x^2 - x +3) - ( x^3 - x^2 +1) . ( 2x - 1 )
a) \(x^3+2x^2-4x+1\)
\(=\left(x^3+3x^2-x\right)-\left(x^2+3x-1\right)\)
\(=x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x-1\right)\left(x^2+3x-1\right)\)
c) cho da thuc P(x) =2x^4-7x^3 -2x^2 +13x +6? | Yahoo Hỏi & Đáp
Tham khảo
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)
\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)
\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)
\(\Leftrightarrow19x-21=-27\)
=>19x=-6
hay x=-6/19
b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)
\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)
\(\Leftrightarrow6x^2=\dfrac{3}{2}\)
\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)
\(\Leftrightarrow3x^2+22x+16=0\)
\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)
\(\Leftrightarrow10x^2-19x=0\)
=>x(10x-19)=0
=>x=0 hoặc x=19/10