Tìm x, biết.
𝑎) (𝑥−1)3+3𝑥(𝑥−4)+1=0
𝑏) (𝑥−1)(𝑥2+𝑥+1)=𝑥2(𝑥−9)+2𝑥2+6
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5 tháng 10 2021 lúc 8:46
BÀI 3: Tìm x, biết.
𝑎) (𝑥−1)3+3𝑥(𝑥−4)+1=0
𝑏) (𝑥−1)(𝑥2+𝑥+1)=𝑥2(𝑥−9)+2𝑥2+6
a) \(\Rightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)
\(\Rightarrow x^3-9x=0\)
\(\Rightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x^3-1=x^3-9x^2+2x^2+6\)
\(\Rightarrow7x^2=7\)
\(\Rightarrow x^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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5 tháng 10 2021 lúc 9:05
Tìm x
𝑎) (𝑥−1)3+3𝑥(𝑥−4)+1=0
𝑏) (𝑥−1)(𝑥2+𝑥+1)=𝑥2(𝑥−9)+2𝑥2+6
Giúp mình với mấy bạn ơi:(
\(a,\Rightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\\ \Rightarrow x^3-9x=0\\ \Rightarrow x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow x^3-1=x^3-9x^2+2x^2+6\\ \Rightarrow7x^2=7\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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a) (𝑥2+1)(𝑥−3)−(𝑥−3)(𝑥2+3𝑥+9)b) (𝑥+2)2+𝑥(𝑥+5)c) (5𝑥+4𝑦)(5𝑥−4𝑦)−24𝑥2+15𝑦2
Đọc tiếp
a) (𝑥2+1)(𝑥−3)−(𝑥−3)(𝑥2+3𝑥+9)
b) (𝑥+2)2+𝑥(𝑥+5)
c) (5𝑥+4𝑦)(5𝑥−4𝑦)−24𝑥2+15𝑦2
a, (x2+1)(x-3)-(x-3)(x2+3x+9)
=(x-3)(x2+1+x2+3x+9)
(x-3)(2x2+3x+10)
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a) (𝑥2+1)(𝑥−3)−(𝑥−3)(𝑥2+3𝑥+9)b) (𝑥+2)2+𝑥(𝑥+5)c) (5𝑥+4𝑦)(5𝑥−4𝑦)−24𝑥2+15𝑦2
Đọc tiếp
a) (𝑥2+1)(𝑥−3)−(𝑥−3)(𝑥2+3𝑥+9)b) (𝑥+2)2+𝑥(𝑥+5)c) (5𝑥+4𝑦)(5𝑥−4𝑦)−24𝑥2+15𝑦2
a) (𝑥2+1)(𝑥−3)−(𝑥−3)(𝑥2+3𝑥+9)b) (𝑥+2)2+𝑥(𝑥+5)c) (5𝑥+4𝑦)(5𝑥−4𝑦)−24𝑥2+15𝑦2
Đọc tiếp
a) (𝑥2+1)(𝑥−3)−(𝑥−3)(𝑥2+3𝑥+9)
b) (𝑥+2)2+𝑥(𝑥+5)
c) (5𝑥+4𝑦)(5𝑥−4𝑦)−24𝑥2+15𝑦2
a) \(\left(x^2+1\right)\left(x-3\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left[x^2+1-\left(x^2+3x+9\right)\right]\)
\(=\left(x-3\right)\left(x^2+1-x^2-3x-9\right)\)
\(=\left(x-3\right)\left(-3x-8\right)\)
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b) \(\left(x+2\right)^2+x\left(x+5\right)\)
\(=x^2+4x+4+x^2+5x\)
\(=2x^2+9x+4\)
c) \(\left(5x+4y\right)\left(5x-4y\right)-24x^2+15y^2\)
\(=25x^2-16y^2-24x^2+15y^2\)
\(=x^2-y^2\)
\(=\left(x+y\right)\left(x-y\right)\)
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1) Làm tính nhâna) 𝑥.(𝑥2–5)b) 3𝑥𝑦(𝑥2−2𝑥2𝑦+3)c) (2𝑥−6)(3𝑥+6)d) (𝑥+3𝑦)(𝑥2−𝑥𝑦)2)Tính (áp dụng Hằng đẳng thức)a) (2𝑥+5)(2𝑥−5)b) (𝑥−3)^2c) (4+3𝑥)^2d) (𝑥−2𝑦)^3e) (5𝑥+3𝑦)^3f) (5−𝑥)(25+5𝑥+𝑥^2)g) (2𝑦+𝑥)(4𝑦^2−2𝑥𝑦+𝑥^2)3)Phân tích các đa thức sau thành nhân tửa) 𝑥^2+2𝑥b) 𝑥^2−6𝑥+9c) 5(𝑥–𝑦)–𝑦(𝑦–𝑥)d) 2𝑥−𝑦^2+2𝑥𝑦−𝑦a) 6𝑥^3𝑦^4+12𝑥^2𝑦^3−18𝑥^3𝑦^2
Đọc tiếp
1) Làm tính nhân
a) 𝑥.(𝑥2–5)
b) 3𝑥𝑦(𝑥2−2𝑥2𝑦+3)
c) (2𝑥−6)(3𝑥+6)
d) (𝑥+3𝑦)(𝑥2−𝑥𝑦)
2)Tính (áp dụng Hằng đẳng thức)
a) (2𝑥+5)(2𝑥−5)
b) (𝑥−3)^2
c) (4+3𝑥)^2
d) (𝑥−2𝑦)^3
e) (5𝑥+3𝑦)^3
f) (5−𝑥)(25+5𝑥+𝑥^2)
g) (2𝑦+𝑥)(4𝑦^2−2𝑥𝑦+𝑥^2)
3)Phân tích các đa thức sau thành nhân tử
a) 𝑥^2+2𝑥
b) 𝑥^2−6𝑥+9
c) 5(𝑥–𝑦)–𝑦(𝑦–𝑥)
d) 2𝑥−𝑦^2+2𝑥𝑦−𝑦
a) 6𝑥^3𝑦^4+12𝑥^2𝑦^3−18𝑥^3𝑦^2
\(1,\\ a,=x^3-5x\\ b,=3x^3y-6x^3y^2+9xy\\ c,=6x^2-6x-36\\ d,=x^3+2x^2y-3xy^2\\ 2,\\ a,=4x^2-25\\ b,=x^2-6x+9\\ c,=9x^2+24x+16\\ d,=x^3-6x^2y+12xy^2-8y^3\\ e,=125x^3+225x^2y+135xy^2+27y^3\\ f,=125-x^3\)
\(g,=8y^3+x^3\\ 3,\\ a,=x\left(x+2\right)\\ b,=\left(x-3\right)^2\\ c,=\left(x-y\right)\left(y+5\right)\\ d,=2x\left(y+1\right)-y\left(y+1\right)=\left(2x-y\right)\left(y+1\right)\\ e,=6x^2y^2\left(xy^2+2y-3x\right)\)
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Tìm mệnh đềphủđịnh mệnh đề𝐴:"∀𝑥∈𝑅,𝑥2−3𝑥=5".
A. 𝐴:"∃𝑥∈𝑅,𝑥2−3𝑥>5".
B. 𝐴:"∃𝑥∈𝑅,𝑥2−3𝑥≠5".
C. 𝐴:"∃𝑥∈𝑅,𝑥2−3𝑥<5".
D. 𝐴:"∃𝑥∉𝑅,𝑥2−3𝑥=5".
(𝑥−1)3−(𝑥+1)(𝑥2−𝑥+1)−(3𝑥+1)(1−3𝑥)
\(\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\\ =\left(x^3-3x^2+3x-1\right)-\left(x^3+1\right)-\left(1-9x^2\right)\\ =x^3-3x^2+3x-1-x^3-1-1+9x^2\\ =6x^2+3x-3\)
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a) 2𝑥(𝑥2−9)=0
b) 2𝑥(𝑥−2021)−𝑥+2021=0
c) 4𝑥2−16𝑥=0
d) (3𝑥+7)2−(𝑥+1)2=0
\(a,\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-1\right)\left(x-2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2021\end{matrix}\right.\\ c,\Leftrightarrow4x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ d,\Leftrightarrow\left(3x+7-x-1\right)\left(3x+7+x+1\right)=0\\ \Leftrightarrow\left(2x+6\right)\left(4x+8\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
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