10000000pp

Ta có: \(x+2x+3x+\cdots+2010x+2011x=2012\cdot2013\)

=>\(x\left(1+2+3+\cdots+2011\right)=2012\cdot2013\)

=>\(x\cdot2011\cdot\frac{\left(2011+1\right)}{2}=2012\cdot2013\)

=>\(x\cdot\frac{2011}{2}\cdot2012=2012\cdot2013\)

=>\(x=2013:\frac{2011}{2}=2013\cdot\frac{2}{2011}=\frac{4026}{2011}\)

a) 2x-7=11x+11

<=> 2x-11x=11+7

<=> -9x=17

<=> x= -17/9

b) 2011x -4 =x+6

<=> 2011x-x=6+4

<=> 2010x=10

<=> x=10/2010

<=> x=1/201

c) 5(2x-3)-2(3x-5)=0

<=> 10x-15-6x+10=0

<=> 10x-6x=15-10

<=>4x=5

<=> x=5/4

Bình phương trong căn phải để trong giá trị tuyệt đối. Tại sao lại bỏ được dấu giá trị tuyệt đối cậu nhỉ?

Vì \(\sqrt{x^2-2x+4} \)≥ 0 ( đúng với ∀ x )
→ \(2x - 2\) ≥ 0 
→x ≥ 1
Ta có : \(\sqrt{x^2-2x+4} \) = \(2x - 2\)
⇔ \(x^2-2x+4 \) = \((2x - 2)^2\)
⇔ \(x^2-2x+4 \) = \(4x^2 - 8x + 4 \)
⇔ \(0 = 3x^2 - 6x \)
⇔ 0 = \(3x(x-1)\)
⇔\(\begin{cases} x=0\\ x-1=0 \end{cases} \)
Mà x ≥ 1
Vậy x ∈ { 1}

Xin lỗi mình lm sai chút :)))
Vì \(\sqrt{x^2-2x+4} \)≥ 0 ( đúng với ∀ x )
→  ≥ 0 
→x ≥ 1
Ta có : \(\sqrt{x^2-2x+4} \) = 
⇔ \(x^2 - 2x + 4\)= 
⇔ 
⇔ 0 = 
⇔\(\left[\begin{array}{} x=0\\ x=2 \end{array} \right.\)
Mà x ≥ 1
→ x ∈ {2}

ĐKXĐ: \(-\dfrac{1}{2}\le x\le\dfrac{1}{2}\)

\(\sqrt{1-2x}+\sqrt{1+2x}=2-x^2\)

\(\Leftrightarrow2+2\sqrt{1-4x^2}=\left(2-x^2\right)^2\)

Đặt \(\sqrt{1-4x^2}=t\ge0\Rightarrow x^2=\dfrac{1-t^2}{4}\)

Pt trở thành:

\(2+2t=\left(2-\dfrac{1-t^2}{4}\right)^2\)

\(\Leftrightarrow\left(t^2+7\right)^2=32\left(t+1\right)\)

\(\Leftrightarrow t^4+14t^2-32t+17=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t+17\right)=0\)

\(\Leftrightarrow t=1\Rightarrow\sqrt{1-4x^2}=1\Rightarrow x=0\)

ĐKXĐ: -1/2<=x<=3/2

Đặt \(a=\sqrt{2x+1}+\sqrt{3-2x}\)

=>\(a^2=2x+1+3-2x+2\cdot\sqrt{\left(2x+1\right)\left(3-2x\right)}=4+2\cdot\sqrt{\left(2x+1\right)\left(3-2x\right)}\)

=>\(a^2=4+2\cdot\sqrt{-4x^2+4x+3}\) >=4

=>a>=2

Theo định lí Cauchy-Schwarz, ta có:

\(a\le\sqrt{\left(1^2+1^2\right)\left(2x+1+3-2x\right)}=2\sqrt2\)

=>\(2\le a\le2\sqrt2\)

\(\left(2x-1\right)^2=4x^2-4x+1=4-\left(-4x^2+4x+3\right)\)

Do đó, ta có:

\(\sqrt{-4x^2+4x+3}=\frac{t^2-4}{2}\)

=>\(-4x^2+4x+3=\frac{\left(t^2-4\right)^2}{4}\)

=>\(\left(2x-1\right)^2=4-\frac{\left(t^2-4\right)^2}{4}=\frac{16-\left(t^2-4\right)^2}{4}\)

=>\(\frac12\left(2x-1\right)^2=\frac{16-\left(t^2-4\right)^2}{8}\)

=>\(t=\frac{16-\left(t^2-4\right)^2}{8}\)

=>\(16-\left(t^2-4\right)^2=8t\)

=>\(16-\left(t^4-8t^2+16\right)=8t\)

=>\(-t^4+8t^2-8t=0\)

=>\(t\left(t^3-8t+8\right)=0\)

=>\(t^3-8t+8=0\)

=>(t-2)\(\left(t^2+2t-4\right)=0\)

TH1: \(t^2+2t-4=0\)

=>\(t^2+2t+1-5=0\)

=>\(\left(t+1\right)^2=5\)

=>\(\left[\begin{array}{l}t+1=\sqrt5\\ t+1=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}t=\sqrt5-1\left(loại\right)\\ t=-\sqrt5-1\left(loại\right)\end{array}\right.\)

TH2: t-2=0

=>t=2(nhận)

=>\(\sqrt{2x+1}+\sqrt{3-2x}=2\)

=>\(2x+1+3-2x+2\cdot\sqrt{\left(2x+1\right)\left(3-2x\right)}=4\)

=>\(2\cdot\sqrt{\left(2x+1\right)\left(3-2x\right)}=0\)

=>(2x+1)(3-2x)=0

=>x=-1/2(nhận) hoặc x=3/2(nhận)

a: Ta có: \(\sqrt{2x-1}=\sqrt{5}\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

hay x=3

b: Ta có: \(\sqrt{9x^2}=2x+1\)

\(\Leftrightarrow\left|3x\right|=2x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\left(x\ge0\right)\\-3x=2x+1\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{-1}{5}\left(nhận\right)\end{matrix}\right.\)