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Bài 1:
a) \(2x^2y-xy=xy\left(2x-1\right)\)
b)\(2x^2-x-2y^2-y=\left(2x^2-2y^2\right)-\left(x+y\right)\)
\(=2\left(x^2-y^2\right)-\left(x+y\right)\)
\(=2\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(2x-2y-1\right)\)
Bài 2:
a)\(x^3-\frac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)=0\)
\(\Rightarrow x=0\text{ hoặc }x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\text{ hoặc }x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{3}\)
Vậy...
b)\(\left(x+1\right)^2=5x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-5x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-5x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(-4x+1\right)=0\)
\(\Leftrightarrow-\left(x+1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\4x=1\Leftrightarrow x=\frac{1}{4}\end{cases}}}\)
Vậy...
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ʌl
Câu 3:
1)
a) Ta có: 3x−2=2x−33x−2=2x−3
⇔3x−2−2x+3=0⇔3x−2−2x+3=0
⇔x+1=0⇔x+1=0
hay x=-1
Vậy: x=-1
b) Ta có: 3−4y+24+6y=y+27+3y3−4y+24+6y=y+27+3y
⇔27+2y=27+4y⇔27+2y=27+4y
⇔27+2y−27−4y=0⇔27+2y−27−4y=0
⇔−2y=0⇔−2y=0
hay y=0
Vậy: y=0
c) Ta có: 7−2x=22−3x7−2x=22−3x
⇔7−2x−22+3x=0⇔7−2x−22+3x=0
⇔−15+x=0⇔−15+x=0
hay x=15
Vậy: x=15
d) Ta có: 8x−3=5x+128x−3=5x+12
⇔8x−3−5x−12=0⇔8x−3−5x−12=0
⇔3x−15=0⇔3x−15=0
⇔3(x−5)=0⇔3(x−5)=0
Vì 3≠0
nên x-5=0
hay x=5
Vậy: x=5
a) 3x - 2 = 2x - 3
\(\Leftrightarrow\) 3x - 2 - 2x + 3 = 0
\(\Leftrightarrow\) x + 1 = 0
\(\Rightarrow\) x = -1
b) 3 - 4y + 24 + 6y = y + 27 + 3y
\(\Leftrightarrow\) 3 - 4y + 24 + 6y - y - 27 - 3y = 0
\(\Leftrightarrow\) -2y = 0
\(\Rightarrow\) y = 0
c)7 - 2x = 22 - 3x
\(\Leftrightarrow\) 7 - 2x - 22 + 3x = 0
\(\Leftrightarrow\) -15 + x = 0
\(\Rightarrow\) x = 15
d) 8x - 3 = 5x + 12
\(\Leftrightarrow\) 8x - 3 - 5x - 12 = 0
\(\Leftrightarrow\)3x -15 = 0
\(\Leftrightarrow\) 3x = 15
\(\Rightarrow\) x = 5
e) x - 12 + 4x = 25 + 2x - 1
\(\Leftrightarrow\) x - 12 + 4x - 25 - 2x + 1 = 0
\(\Leftrightarrow\) 3x - 36 = 0
\(\Leftrightarrow\) 3x = 36
\(\Rightarrow\) x = 12
f ) x + 2x + 3x - 19 = 3x + 5
\(\Leftrightarrow\) x + 2x + 3x - 19 - 3x - 5 = 0
\(\Leftrightarrow\)3x - 24 = 0
\(\Leftrightarrow\) 3x = 24
\(\Rightarrow\) x = 8
g) 11+ 8x - 3 = 5x - 3 +x
\(\Leftrightarrow\)8x + 8 = 6x - 3
\(\Leftrightarrow\)8x - 6x = -3 - 8
\(\Leftrightarrow\)2x = -11
\(\Rightarrow\)x = \(-\frac{11}{2}\)
h) 4 - 2x +15 = 9x + 4 -2
\(\Leftrightarrow\)19 - 2x = 7x + 4
\(\Leftrightarrow\)-2x - 7x = 4 - 19
\(\Leftrightarrow\)-9x = -15
\(\Rightarrow\)x = \(\frac{15}{9}\) = \(\frac{5}{3}\)
\(A=''2^9+2^7+1''''2^{23}-2^{21}+2^{19}-2^{17}+2^{14}-2^{10}+2^9-2^7+1''\)
Thực hiện phép tính đầu
\(2^9=2\times2\times2\times2\times2\times2\times2\times2\times2=512\)
\(2^7=2\times2\times2\times2\times2\times2\times2=128\)
\(=128+512+1=641\)
Phép tính thứ hai là tương tự như phép tính thứ nhất
Nhân lên rồi cộng vào nha
Đặt cột dọc ra chia :v
Sửa đề: \(\left(3x^3-2x^2+2x+1\right):\left(3x+1\right)\)
\(=\left(3x^3-3x^2+3x+x^2-x+1\right):\left(3x+1\right)\)
\(=\left[3x\left(x^2-x+1\right)+\left(x^2-x+1\right)\right]:\left(3x+1\right)\)
\(=\left[\left(x^2-x+1\right)\left(3x+1\right)\right]:\left(3x+1\right)\)
\(=x^2-x+1\)