x2 – 3x + 7 = 1 + 2x
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30 tháng 9 2021
\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)
PN
22 tháng 5 2021
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
PN
22 tháng 5 2021
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
12 tháng 9 2021
\(a,=6x^2+23x+21-\left(6x^2+23x-55\right)\\ =76\left(đpcm\right)\\ b,=3x^4+6x^3+9x^2-2x^3-4x^2-6x+x^2+2x+3-4x^3+4x-3x^4-6x^2\\ =3\left(đpcm\right)\)
19 tháng 9 2023
a)-(x-y)(x2+xy-1)=-(x3+x2y-x-x2y-xy2+y)
=-(x3-xy2-x+y)
=-x3+xy2+x-y
b)x2(x-1)-(x3+1)(x-y)=x3-x2-x3+x2y-x+y
=-x2+x2y-x+y
c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x2-3x-4x+2-15x2-10x-3x-2
=-9x2-20x
d) hình như bạn ghi lỗi
Bài 2: C=x(x2-y)-x2(x+y)+y(x2-x)
=x3-xy-x3-x2y+x2y-xy
=-2xy
Thay x=1/2,y=-1 vào C, ta có:
C=-2.1/2.(-1)=1
Vậy C=1 khi x=1/2 và y=-1.
3 tháng 2 2022
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
Vì \(x^2+2\ge2>0\forall x\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)
TL
0
10 tháng 1 2024
a: \(x+7⋮x+2\)
=>\(x+2+5⋮x+2\)
=>\(5⋮x+2\)
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
b: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
c: \(3x-2⋮x+3\)
=>\(3x+9-11⋮x+3\)
=>\(-11⋮x+3\)
=>\(x+3\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-2;-4;8;-14\right\}\)
d: \(12x+1⋮3x+2\)
=>\(12x+8-7⋮3x+2\)
=>\(-7⋮3x+2\)
=>\(3x+2\in\left\{1;-1;7;-7\right\}\)
=>\(3x\in\left\{-1;-3;5;-9\right\}\)
=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{5}{3};-3\right\}\)
e: \(x^2+3x+5⋮x+3\)
=>\(x\left(x+3\right)+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
f: \(x^2-2x+3⋮x+2\)
=>\(x^2+2x-4x-8+11⋮x+2\)
=>\(11⋮x+2\)
=>\(x+2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-1;-3;9;-13\right\}\)
\(2x-3x+7=1+2x\)
\(2x-3x-2x=1-7\)
\(-3x=-6\)
\(x=2\)
x^2 – 3x + 7 = 1 + 2x
=> x^2 - 5x + 6 = 0
=> x^2 - 2x - 3x + 6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 2)(x - 3) = 0
=> x = 2 hoặc x = 3