Tìm x biết
1 + 2 + 3 + 4 + ..... + x = aaa
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1)
\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(3x-6+4x-4=25\)
\(7x-10=25\\ 7x=35\\ x=5\)
2)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\left(x-2\right)\left(4x-2\right)=0\)
\(=>\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
3)
\(\left(x-2\right)^2=4\left(x-1\right)^2\)
\(x^2-4x+4=4\left(x^2-2x+1\right)\)
\(x^2-4x+4=4x^2-8x+4\)
\(x^2-4x+4-4x^2+8x-4=0\)
\(-3x^2+4x=0\)
\(x\left(-3x+4\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\-3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2
3)
(2x+1)2=52
2x+1=5
2x=4
x=2
\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)
\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)
Tìm các số a,b,c biết
1,-3x^3.(2ax^2-bx+c)=-6x^5+9x^4-3x^3
2,(ax+b)(x^2-cx+2)=x^3+x^2-2
Giúp mình với ạ
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
1. x(x + 1) - x2 + 1 = 0
<=> x(x + 1) - (x2 - 1) = 0
<=> x(x + 1) - (x + 1)(x - 1) = 0
<=> (x - x + 1)(x + 1) = 0
<=> x + 1 = 0\
<=> x = -1
2. 4x(x - 2) - 6 + 3x = 0
<=> 4x(x - 2) - (3x - 6) = 0
<=> 4x(x - 2) - 3(x - 2) = 0
<=> (4x - 3)(x - 2) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)
3. x(x + 2) - 3(x + 2) = 0
<=> (x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Bài 10:
a: Để A là phân số thì n+2<>0
hay n<>-2
b: Khi n=0 thì A=3/2
Khi n=2 thì A=3/(2+2)=3/4
Khi n=-7 thì A=3/(-7+2)=-3/5
Bài 9:
1)9/x = -35/105 2) 12/5 = 32/x 3)x/2 = 32/x x = 9. (-35)/105 x.12/5 = x.32/x 2x.x/2 = 2x.32/x
x = -3 x.12/5=32 xx = 2.32
x= 32:12/5 x^2 = 2.32
x = 40/3 x^2 = 64
x = 8
4) x-2/4 = x-1/5
5(x-2) = 4(x-1)
5x - 10 = 4x - 4
5x - 4x = 10 - 4
x = 6
Bài 10:Cho biểu thức A=3/n+2
a) Để A là phân số thì mẫu số phải khác 0
Do đó: n + 2 ≉ 0. Suy ra: n ≉ -2
b) Khi n = 0 thì A = 3/0+2 = 3/2
Khi n = 2 thì A = 3/2+2 = 3/4
Khi n = -7 thì A = 3/-7+2 = 3/-5