\(b)\) \(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\)\(\left(2x-1\right)^{2010}.\left(2x-1\right)^2=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2=1\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=0\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{2}\\x=\frac{0}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

Vậy \(x=0\) hoặc \(x=1\)

Chúc bạn học tốt ~ 

a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy phương trình có nghiệm x = - 5 .

a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)

\(\Rightarrow3x^2-3x-3x^2+2x=5\)

\(\Rightarrow5x=5\Rightarrow x=1\)

Câu b,c làm tương tự! Cứ tách ra là làm được à!

b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)

\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)

\(\Leftrightarrow20x=-40\)

\(\Leftrightarrow x=-2\)

Vậy phương trình có nghiệm x = - 2

x=0

ban

a, (2x-3)⁴=(2x-3)⁶

=> (2x-3)⁶ : (2x-3)⁴=1

=> (2x-3)³=

=> 2x-3=1

=> 2x=4

=> x=2

b, (3x+5)³=(3x+5)²⁰¹⁶

=> (3x+5)²⁰¹⁶ : (3x+5)³=1

=> (3x+5)²⁰¹³=1

=> 3x+5=1

=> 3x=-4

=> x=-4/3

c, (2x+1)²⁰¹⁵=(2x+1)²⁰¹⁷

=> (2x+1)²⁰¹⁷ : (2x+1)²⁰¹⁵=1

=> (2x+1)²=1

=> 2x+1=1

=> 2x=0

=> x=0

1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=15/93

1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3)=15/93

1/2(1/3-1/2x+3)=15/93

=>1/3-1/2x+3=10/31

=>1/2x+3=1/93

=>2x+3=93

2x=93-3=90

=>x=45

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(x=45\)

Vậy \(x=45\).

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

=> \(\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{5}{31}\)

=> \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{5}{31}\)

=> \(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{31}:\frac{1}{2}=\frac{5}{31}\cdot2=\frac{10}{31}\)

=> \(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)

=> 2x +3  = 93

=> 2x = 90 => x = 45

(x - 3)² = (2x - 1)²

=> |x - 3| = |2x - 1|

=> x - 3 = 2x-1 hoặc x - 3 = -(2x - 1)

x - 2x = -1 + 3 hoặc x - 3 = -2x + 1

-x = 2 hoặc x + 2x = 1 + 3

x = -2 hoặc 3x = 4

x = -2 hoặc x = 4 : 3

x = -2 hoặc x = 1,(3)

Bạn xem lại đề nhé.

a) \(A=x^2+5y^2+2xy-4x-8y+2015\)

\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2-y\right)^2+4y^2+2011\)

Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)

\(\Rightarrow A_{min}=2011\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)

\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)

\(B=2x^2+2x+2012^2+2013^2\)

\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)

\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)

\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)

Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)

a) \(36x^2-12x-36x^2+27x=30\)

                                                \(15x=30\)

                                                     \(x=2\)

b) \(5x-2x^2+2x^2-2x=15\)

                                          \(3x=15\)

                                             \(x=5\)

Đề bạn sai câu b/

Ta có:

|2x-5|+1=3x

=> |2x-5|=3x-1

=> 2x-5=3x-1 ; 2x-5=1-3x

=> -5+1=3x-2x ; 2x+3x=1+5

=>x=-4 ; 5x =6

; x=\(\dfrac{5}{6}\)