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VN
0
TS
17 tháng 1 2018
Thực ra 2 câu đầu rất dễ nha bạn ^^!
1) x4 + 2x3 + x2 + 2x + 1 =0 <=> x3(x+2)+x(x+2)+1 = 0
<=> (x3+x)(x+2) + 1=0
1>0
=> (x3+x)(x+2) + 1=0 <=> (x3+x)(x+2) = 0
<=>\(\orbr{\begin{cases}^{x^3+x=0}\\x+2=0\end{cases}}\)<=>\(\orbr{\begin{cases}^{x\left(x^2+1\right)=0}\\x=-2\end{cases}}\) <=>
\(\orbr{\begin{cases}^{x=0}\\x=-2\end{cases}}\)
b)
x3+1=\(2\sqrt[3]{2x-1}\)
<=> x^3 - 1 = 2(\(\sqrt[3]{2x-1}\) -1)
<=> (x-1)(x2+x+1) = \(\frac{4\left(x-1\right)}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\)
<=> (x-1)[(x2+x+1) - \(\frac{1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\) ] =0
<=> x=1
H
28 tháng 7 2023
\(\left(x+2\right)^2=\left(2x-1\right)^2\\ \Leftrightarrow\left(x+2\right)^2-\left(2x-1\right)^2=0\\\Leftrightarrow\left[x+2-\left(2x-1\right)\right]\left[x+2+2x-1\right]=0\\ \Leftrightarrow\left(x+2-2x+1\right)\left(x+2+2x-1\right)=0\\ \Leftrightarrow\left(-x+3\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
28 tháng 7 2023
\(\left(x+2\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-1\\x+2=-\left(2x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-2\\x+2=-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\x+2x=1-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
27 tháng 6 2021
\(a,=x^2-1-\left(x^2+4x+4\right)=x^2-1-x^2-4x-4=11\)
\(\Leftrightarrow-5x=15\)
\(\Leftrightarrow x=-3\)
Vậy ...
\(b,=\left(x-3-2x+5\right)\left(x-3+2x-5\right)=0\)
\(\Leftrightarrow\left(-x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy ...
TG
27 tháng 6 2021
a) \(\left(x-1\right)\left(x+1\right)-\left(x+2\right)^2=11\)
\(\Rightarrow x^2-1-x^2-4x-4-11=0\)
=> -4x - 16 = 0
=> -4x = 16
=> x = -4
b) \(\left(x-3\right)^2-\left(2x-5\right)^2=0\)
=> (x - 3 + 2x - 5).(x - 3 - 2x + 5) = 0
=> (3x - 8).(-x + 2) = 0
=> x = 8/3 hoặc x = 2
C
2
5 tháng 2 2018
\(\left(2x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)
\(2\left(x+2\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)
\(\left(x+2\right)\left(2x-6-x+4\right)=x\left(x+5\right)\)
\(\left(x+2\right)\left(x-2\right)-x^2-5x=0\)
\(x^2-2x+2x-4-x^2-5x=0\)
\(-5x-4=0\)
\(-5x=4\)
\(\Rightarrow\)\(x=\frac{-4}{5}\)
\(\left(x-2\right)^2=\left(2x-4\right)\left(x+5\right)\)
\(\left(x-2\right)^2-2\left(x-2\right)\left(x+5\right)=0\)
\(\left(x-2\right)\left(x-2-2x-10\right)=0\)
\(\left(x-2\right)\left(-x-12\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\-x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}}\)
Bạn tự kết luận 2 câu nhé
b) 
d) 
f) 
h) 
k) 
n) 
1 tháng 2 2022
bạn đăng tách cho mn cùng giúp nhé
Bài 1 :
a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)
c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)
d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)
e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)
f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)
28 tháng 12 2016
\(\left(2+1\right)\left(2^2+1\right)...\left(2^8+1\right)-2^{16}=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^8+1\right)-2^{16}\)\(2^{16}\)
\(=-1\)
PN
20 tháng 11 2015
Nhớ ghi dấu ngoặc tránh giải sai.
\(a.\) \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}\)
Ta có:
\(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
nên \(MTC:\) \(2\left(x-3\right)\left(x+3\right)\)
Do đó: \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}=\frac{x+4}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+4\right)\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}+\frac{2.3}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2+x-12+6}{2\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2-2x+3x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-2\right)+3\left(x-2\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{x-2}{2\left(x-3\right)}\)
(x-2)(x2 + 2x+4)
= x3 - 23
= x3 -8