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\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Bài 1.
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
a: \(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a) \(x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)
e) \(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)
h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
P=n3+4n-5=n3-n+5n-5=n(n2-1)+5(n-1)
=n(n-1)(n+1)+5(n-1)=(n-1)[n(n+1)+5]
=(n-1)(n2+n+5)
Vì n \(\in\) N nên n2+n+5 > 1
Để P là số nguyên tố thì n-1=1=>n=2
Thử lại thấy n=2 thỏa mãn
Vậy n=2
1) a) x = -7 / 44
b) x = -1 / 8