(x mũ 2 + 4)mũ 2 -16x mũ 2
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MD
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
\(A=x^3-15x^4+16x^3-29x^2\)
\(A=\left(x^3+16x^3\right)-15x^4-29x^2\)
\(A=17x^3-15x^4-29x^2\)
\(A=-15x^4+17x^3-29x^2\)
\(A=-x^2\left(15x^2-17x+29\right)\)
TN
1
24 tháng 4 2023
a: =>9(2x+1)=6(3-x)
=>3(2x+1)=2(3-x)
=>6x+3=6-2x
=>8x=3
=>x=3/8
b: =>-3x^2-2+3x^2-18x=-26
=>-18x=-24
=>x=4/3
ND
26 tháng 5 2018
1. \(x^4-2x^2+1=\left(x^2-1\right)^2\)
2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
3. \(16x^2-8x+1=\left(4x-1\right)^2\)
4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)
5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)
6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)
7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)
8 tháng 8 2021
6, \(x^2-1+2xy+y^2=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)
8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)
9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)
10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)
8 tháng 8 2021
7, 4x mũ 2 - 12x + 9 - y mũ 2 = -(y-2x+3) (y+2x-3)
8, 16x mũ 2 - 4y mũ 2 + 4y - 1 = -(2y - 4x - 1) (2y+4x-1)
9, 25 - x mũ 2 - 12x - 36 = -(x+1) (x+11)
10, x mũ 2 - 9 - 5 ( x + 3 ) = (x-8) (x+3)
bạn k cho mình nha
chúc bạn học tốt :))))
8 tháng 8 2021
bạn kham khảo link, mình đã làm rồi nhé
Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM
ND
2 tháng 6 2018
1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)
2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-5x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(x-5\right)^2=0\Leftrightarrow x=5\)
5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)
7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)
\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)
\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
PT
1
\(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2+4-4x\right).\left(x^2+4+4x\right)\)
\(=\left(x-2\right)^2.\left(x+2\right)^2=\left(x^2-4\right)^2\)
16x mũ 2 trừ y mũ 2 mũ 2