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FC
14 tháng 8 2019
\(a,\) \(3a^3.\left(x^2-1\right)^4:3a^3.\left(x^2-1\right)^3=15\)
\(\frac{3a^3.\left(x^2-1\right)^4}{3a^3.\left(x^2-1\right)^3}=15\\ x^2-1=15\\ x^2=16\\ x=4\)
\(b,\) \(x^3.\left(2x-1\right)^{m+2}:x^3.\left(2x-1\right)^{m-1}=3^5:3^2\\ \frac{x^3.\left(2x-1\right)^{m+2}}{x^3\left(2x-1\right)^{m-1}}=3^{5-2}\\ \left(2x-1\right)^3=3^3\)
\(2x-1=3\\ 2x=4\\ x=2\)
13 tháng 7 2018
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3
NV
27 tháng 7 2017
a)(3x-1)2+2(3x-1)(2x+1)2(2x+1)=48x^4+56x^3+21x^2-12x-1 cái này tra google
b)(x2+1)(x-3)-(x-3)(x2+3x+9)=(x2+1)(x-3)-(x-3)(x+3)2=(x-3)[(x2+1)-(x+3)2 ]
c)(2x+3)2+(2x+5)2-2(2x+3)(2x+5)=(2x+3)2+(2x+5)2-(2x+3)(2x+5)-(2x+3)(2x+5)=(2x+3)(2x+3-2x+5)+(2x+5)(2x+5-2x+3)
=8(2x+3)+8(2x+5)=8(2x+3+2x+5)
=8(4x+8)
d)(x-3)(x+3)-(x-3)2 =(x-3)(x+3)-(x-3)(x-3)=(x-3)(x+3-x-3)=0
e)(2x+1)2+2(4x2-1)+(2x-1)2 =(2x+1)2+2[(2x)2 -1]+(2x-1)2 =(2x+1)(2x+1+2x-1)+(2x-1)(2x+1+2x-1)=4x(2x+1)+4x(2x-1)
=4x(2x+1+2x-1)=16x2
f)(x2-1)(x+2)-(x-2)(x2+2x+4)= (x2-1)(x+2)-(x-2)(x+2)2 =(x2-1)(x+2)-(x2-22)(x+2)=(x+2)(x2-1-x2-22) mình đoán câu f khai triển ra thế này nhưng kq không giống nhau nên chắc bạn phải tự làm rồi
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
13 tháng 1 2017
1. Ta có \(x^3+3x^2+x+3=0\)
\(\Leftrightarrow\left(x^3+3x^2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)
Nếu x+3=0 =>x=-3
Nếu \(x^2+1=0\) =>x\(=\varnothing\) (vì \(x^2+1>0\))
Vậy x=-3
DN
13 tháng 1 2017
2) đặt x^2+x+1 = t
=> x^2 +x +2 =t+1
pt => t(t+1)=2
t^2 + t -2 =0
\(\Rightarrow\left[\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
voi t=1 => x^2 +x+1=1
=> \(\Rightarrow\left[\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
voi t=-2 => x^2+x+1=-2
=> x^2+x+3=0(vo nghiem)
cau 3 lam nhu cau 2
4) pt <=> (x^2-4)(x+3-x+1)=0
ban tu giai not nha
NN
15 tháng 11 2017
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
NN
15 tháng 11 2017
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
14 tháng 8 2019
a) \(3a^3\left(x^2-1\right)^4:3a^3\left(x^2-1\right)^3\)
\(=3a^3\left(x^2-1\right)^3\left(x^2-1\right):3a^3\left(x^2-1\right)^3\)
\(=x^2-1\)
Mà \(3a^3\left(x^2-1\right)^4:3a^3\left(x^2-1\right)^3=15\)
\(\Rightarrow x^2-1=15\)
\(\Leftrightarrow x=\pm4\)
14 tháng 8 2019
b) \(x^3\left(2x-1\right)^{m+2}:x^3\left(2x-1\right)^{m-1}\)
\(=x^3\left(2x-1\right)^{m-1}.m^3:x^3\left(2x-1\right)^{m-1}\)
\(=m^3\)
Mà \(x^3\left(2x-1\right)^{m+2}:x^3\left(2x-1\right)^{m-1}=3^5:3^2\)
\(\Rightarrow m^3=3^5:3^2\)
\(\Leftrightarrow m^3=3^3\)
\(\Leftrightarrow m=3\)
\(\left(x^2-1\right)^3+\left(x^2+2\right)^3+\left(2x-1\right)^3+\left(3x+3\right)\left(2x-1\right)\)\(.\left(1-x\right)\left(x^2+2\right)=0\).
\(\Leftrightarrow\left(x^2-1\right)^3+\left(x^2+2\right)^3+\left(2x-1\right)^3-3\left(x+1\right)\left(2x-1\right)\)\(.\left(x-1\right)\left(x^2+2\right)=0\).
\(\Leftrightarrow\left(x^2-1\right)^3+\left(x^2+2\right)^3+\left(2x-1\right)^3-3\left(x^2-1\right)\left(2x-1\right)\)\(.\left(x^2+2\right)=0\).
Đặt \(x^2-1=a\), \(x^2+2=b\), \(2x-1=c\). Phương trình trở thành:
\(a^3+b^3+c^3-3abc=0\).
\(\Leftrightarrow\left[a^3+b^3+3ab\left(a+b\right)\right]-3ab\left(a+b\right)+c^3-3abc=0\).
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\).
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\).
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\).
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\).
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\).
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\).
Xét phương trình \(\left(1\right)\).
\(\left(1\right)\Leftrightarrow x^2-1+x^2+2+2x-1=0\)
\(\Leftrightarrow2x^2+2x=0\).
\(\Leftrightarrow2x\left(x+1\right)=0\).
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\).
Xét phương trình \(\left(2\right)\)
\(\left(2\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0.2\).
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\).
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\).
\(\Leftrightarrow\left(x^2-1-x^2-2\right)^2+\left(x^2+2-2x+1\right)^2+\left(2x-1-x^2+1\right)^2=0\).
\(\Leftrightarrow\left(-3\right)^2+\left[\left(x-1\right)^2+2\right]^2+\left(-x^2+2x\right)^2=0\).
\(\Leftrightarrow\left[\left(x-1\right)^2+2\right]^2+\left(x^2-2x\right)^2+9=0\).
Ta có:
\(\left(x-1\right)^2\ge0\forall x\).
\(\Rightarrow\left(x-1\right)^2+2\ge2\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2\ge4\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9\ge13\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9>0\forall x\).
\(\left(x^2-2x\right)^2\ge0\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9+\left(x^2-2x\right)^2>0\forall x\).
Do đó phương trình \(\left(2\right)\)vô nghiệm.
Vậy phương trình có tập nghiệm: \(S=\left\{0;-1\right\}\).
\(\)