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Bài 1:
a: x^3-6x^2+12x-7=0
=>x^3-x^2-5x^2+5x+7x-7=0
=>(x-1)(x^2-5x+7)=0
=>x=1
b: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)
=>4x^2-5x-4x^2-4x-1=0
=>-9x-1=0
=>9x+1=0
=>x=-1/9
A = \(\frac{3}{5}\)a - \(\frac{4}{5}\) b2
B = 3x - 5y + 4x
a/ (x+y)3-(x-y)3-2y3
= (x3+3x2y+3xy2+y3)-(x3-3x2y+3xy2-y3)-2y3
= x3+3x2y+3xy2+y3-x3+3x2y-3xy2+y3-2y3
= 6xy2
b/ (x+2)(x2-2x+4)-(16-x3)
= x3-2x2+4x+2x2-4x+8-16+x3
= 2x3-8
c/ (2a+b)(4a2-2ab+b2)-(2a-b)(4a2+2ab+b2)
= (8a3+b3)-(8a3-b3)
= 8a3+b3-8a3+b3
= 2b3
a, \(\dfrac{3a^2b-4ab^2}{5ab}=\dfrac{ab\left(3a-4b\right)}{5ab}=\dfrac{3a-4b}{5}\)
b, \(\dfrac{3x^3y^2-5x^2y^3+4x^3y^3}{x^2y^2}=\dfrac{x^2y^2\left(3x-5y+4xy\right)}{x^2y^2}\)
\(=3x-5y+4xy\)
c, \(\dfrac{2a^5b^4+3a^4b^3}{-3a^4b^5}=\dfrac{a^4b^3\left(2ab+3\right)}{-3a^4b^5}=\dfrac{2ab+3}{-3b^2}\)
d, \(\dfrac{-a^5b^4+3a^6b^2}{4a^4b^2}=\dfrac{-a^4b^2\left(ab^2+3a^2\right)}{4a^4b^2}=\dfrac{-\left(ab^2+3a^2\right)}{4}\)
Chúc bạn học tốt!!!
a. \(\left(3a^2b-4ab^3\right):5ab=3a^2b:5ab-4ab^3:5ab=\dfrac{3}{5}a-\dfrac{4}{5}b^2\)
b. \(\left(3x^3y^2-5x^2y^3+4x^3y^3\right):x^2y^2=3x^3y^2:x^2y^2-5x^2y^3:x^2y^2+4x^3y^3:x^2y^2=3x-5y+4xy\)
c. \(\left(2a^5b^4+3a^4b^3\right):\left(-3a^4b^5\right)=2a^5b^4:\left(-3a^4b^5\right)+3a^4b^3:\left(-3a^4b^5\right)=-\dfrac{2a}{3b}-\dfrac{1}{b^2}\)
d. \(\left(-a^5b^4+3a^6b^2\right):4a^4b^2=\left(-a^5b^4\right):4a^4b^2+3a^6b^2:4a^4b^2=-\dfrac{1ab^2}{4}+\dfrac{3a^2}{4}\)
bn chép lại đề nhé
a/ \(=\left(x+y\right)^2-4x^2y^2=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
b/ \(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[-\left(b+c\right)^2+a^2\right]\)
\(=\left(b+c-a\right)\left(b+c+a\right)^2\left(a-b-c\right)\)
c/ \(=2a^2+2b^2-2c^2+4ab=2\left[\left(a^2+b^2+2ab\right)-c^2\right]\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)
d/ \(=\left(4x^2-25\right)^2-9\left(4x^2-20x+25\right)\)
\(=\left(4x^2-25\right)^2-9\left(4x^2+25\right)+180x\)
tới đây bạn đặt a= 4x^2 -25 rồi làm típ nha, mình lười quá ><
e/ tương tự câu d nha bạn
f/ \(=a^4\left(a^2-1\right)+2a^2\left(a+1\right)\)
\(=a^4\left(a-1\right)\left(a+1\right)+2a^2\left(a+1\right)\)
\(=a^2\left(a+1\right)\left(a^2+2\right)\)
g/ đặt \(a=3x^2+3x+2\) khi đó biểu thức trở thành
\(a^2-\left(a+4\right)^2=a^2-a^2-8a-16\)
\(=-8a-16=-8\left(3x^2+3x+2-8\right)=-8\left(3x^2+3x-6\right)\)
\(=-24\left(x^2+x-2\right)=-24\left(x-1\right)\left(x+2\right)\)
xong rùi nha bn. Chúc bn hc tốt (xin lỗi tại có mấy câu mình lười nha)
\(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2\)
\(=\left(x-2xy+y\right)\left(x+2xy+y\right)\)
Bài 1:
Ta có: \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)
\(=\left(2a-3b\right)^2-2\cdot\left(2a-3b\right)\cdot\left(2b-3a\right)+\left(2b-3a\right)^2\)
\(=\left(2a-3b-2b+3a\right)^2\)
\(=\left(5a-5b\right)^2\)
\(=\left[5\cdot\left(a-b\right)\right]^2=25\left(a-b\right)^2\)
Thay a-b=0 vào biểu thức \(A=25\left(a-b\right)^2\), ta được:
\(A=25\cdot0^2=0\)
Vậy: Khi a-b=0 thì A=0
Bài 3:
a) Ta có: \(A=x^2+8x\)
\(=x^2+8x+16-16\)
\(=\left(x+4\right)^2-16\)
Ta có: \(\left(x+4\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x+4\right)^2-16\ge-16\forall x\)
Dấu '=' xảy ra khi x+4=0
hay x=-4
Vậy: Giá trị nhỏ nhất của biểu thức \(A=x^2+8x\) là -16 khi x=-4
Bài 2:
a, \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy...
b, \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
c, \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
Bài 3:
1, Đặt \(A=x^2+\dfrac{1}{2}x+\dfrac{1}{16}=x^2+\dfrac{1}{4}.x.2+\dfrac{1}{16}\)
\(=\left(x+0,25\right)^2\)
Thay x = 49,75 vào A ta có:
\(A=50^2=2500\)
2, tương tự
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)