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k cho mk nha
x^4-2x^3+3x^2-2x+1
=(x^4-2x^3+x^2)+(x^2-2x+1)
=x^2(x^2-2x+1)+(x^2-2x+1)
=(x^2+1)(x^2-2x+1)
=(x^2+1)(x-1)^2
\(\left(8x^3-7x^2\right)\div x^2=3x+\sqrt{\frac{9}{25}}\)
\(\Leftrightarrow\left(8x^3\div x^2\right)-\left(7x^2\div x^2\right)=3x+\frac{3}{5}\)
\(\Leftrightarrow8x-7=3x+\frac{3}{5}\)
\(\Leftrightarrow8x-3x=\frac{3}{5}+7\)
\(\Leftrightarrow5x=\frac{38}{5}\)
\(\Leftrightarrow x=\frac{38}{25}\)
talaays đơn thức nhân với từng hạng tử của đa thức
rồi cộng tích lại với nhau
rồi tìm x
nha bn
Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
a. (x + 3).(x2 - 1)
= x.x2 - x.1 + 3.x2 - 3.1
= x3 - x + 3x2 - 3
= x3 + 3x2 - x - 3
b. (3x + 2).(4x - 1)
= 3x.4x - 3x + 2.4x - 2
= 12x2 - 3x + 8x - 2
= 12x2 + 5x - 2
c. (2x - 3).(3x + 2)
= 2x.3x + 2x.2 - 3.3x - 3.2
= 6x2 + 4x - 9x - 6
= 6x2 - 5x - 6
d. (12x - 5).(4x + 1)
= 12x.4x + 12x - 5.4x - 5
= 48x2 + 12x - 20x - 5
= 48x2 - 8x - 5
e. (x - 3).(x2 + 3x + 9)
= x.x2 + x.3x + x.9 - 3x2 - 3.3x - 3.9
= x3 + 3x2 + 9x - 3x2 - 9x - 27
= x3 - 27 (Đây là dạng HĐT x3 - 33)
a,\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
Ta có: \(x^2+5\ge0\) (vô lí)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-6\end{cases}}\)
Vậy ....
c, \(4x^2\left(x-1\right)-x+1=0\)
\(\Leftrightarrow4x^3-4x^2-x+1=0\)
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x^2-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x^2=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\frac{1}{2}\\x=1\end{cases}}\)
Vậy ....
\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)
ĐKXĐ: \(x\ne1,x\ne-3\)
PT đã cho \(\Leftrightarrow\frac{\left(x+2\right).\left(x-1\right)-\left(x+1\right).\left(x+3\right)}{\left(x+3\right).\left(x-1\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x-1\right)-\left(x+1\right).\left(x+3\right)}{\left(x+3\right).\left(x-1\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+x-2-x^2-4x-3=4\Leftrightarrow3x=-1\Leftrightarrow x=\frac{-1}{3}\)