\(3x^{2013}+5x^{2011}+2006\) . với \(x=\sqrt{6+2\sqrt{2}.\sq...">
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28 tháng 11 2019

Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen

help me, pleaseee

Cần gấp lắm ạ!

16 tháng 12 2017

a)x=6

b)x=6

d)x=0.2

12 tháng 8 2017

đăng ít một thôi bạn

12 tháng 8 2017

Bỏ câu c,d đi ạ 

1 tháng 7 2019

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

\(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

1 tháng 7 2019

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá

29 tháng 7 2018

1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)

<=> \(\sqrt{\left(x-10\right)^2}=10\)

<=> \(\left|x-10\right|=10\)

=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)

Vậy S = \(\left\{20;0\right\}\)

2) \(\sqrt{x +2\sqrt{x}+1}=6\)

<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)

<=> \(\left|\sqrt{x}+1\right|=6\)

=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)

Vậy S = \(\left\{25\right\}\)

3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)

<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)

<=> \(\left|x-3\right|=\sqrt{3}+1\)

=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)

Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)

29 tháng 7 2018

4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)

<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)

<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)

<=> \(\left|\sqrt{3x}+1\right|=5\)

=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)

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

<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)

<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)

Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)

6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)

<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)

<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)

<=> \(\left|\sqrt{6x}+2\right|=7\)

=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)

=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)

1 tháng 8 2018

1) \(\sqrt{x-1}=\sqrt{2x+3}\) ĐK: x ≥ 1; x ≥ \(\dfrac{-3}{2}\) => x ≥ 1

=> x - 1 = 2x + 3

=> x - 2x = 3 + 1

=> -x = 4 => x = -4 (ko TMĐK)

Vậy S = ∅

2) \(\sqrt{2x-3}=\sqrt{x-1}\) ĐK: x ≥ \(\dfrac{3}{2}\); x ≥ 1 => x ≥ \(\dfrac{3}{2}\)

=> 2x - 3 = x - 1

=> 2x - x = -1 + 3

=> x = -2 (ko TMĐK)

Vậy S = ∅

3) \(\sqrt{2-x}=\sqrt{3+x}\) ĐK: x ≥ 2; x ≥ -3 => x ≥ 2

=> 2 - x = 3 + x

=> -x - x = 3 - 2

=> -2x = 1 => x = \(\dfrac{-1}{2}\) (ko TMĐK)

Vậy S = ∅

4) \(\sqrt{4x-8}=2\sqrt{x-2}\) ĐK: x ≥ 2

=> 4x - 8 = 2(x - 2)

=> 4x - 8 = 2x - 4

=> 4x - 2x = -4 + 8

=> 2x = 4 => x = 4 : 2 = 2 (TMĐK)

Vậy S = \(\left\{2\right\}\)

5) \(\sqrt{x^2-5}=\sqrt{4x-9}\) ĐK: \(\left|x\right|=\sqrt{5}\) ; x ≥ \(\dfrac{9}{4}\)

<=> x2 - 5 = 4x - 9

<=> x2 - 4x - 5 + 9 = 0

<=> x2 - 4x - 4 = 0 <=> (x - 2)2 =0

=> x = 2 (ko TMĐK)

6) \(\sqrt{x-2}=\sqrt{x^2-2x}\) ĐK: x ≥ 2

=> x - 2 = x2 - 2x

=> x - 2 - x2 + 2x = 0

=> (x - 2) - x(x - 2) = 0

=> (1- x) . (x - 2) = 0

=> \(\left\{{}\begin{matrix}1-x=0\\x-2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-0=1\left(loai\right)\\x=0+2=2\left(TMĐK\right)\end{matrix}\right.\)

Vậy S = \(\left\{2\right\}\)

7) \(\sqrt{x^2-3x}-\sqrt{15-5x}=0\) ĐK: x ≥ 3 hoặc x ≤ 0

<=> \(\sqrt{x^2-3x}=\sqrt{15-5x}\)

<=> x2 - 3x = 15 - 5x

=> x2 - 3x + 5x - 15 = 0

=> x(x -3) + 5(x - 3) = 0

=> (x + 5) . (x - 3) = 0

=> \(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=0-5=-5\\x=0+3=3\end{matrix}\right.\)(TMĐK)

Vậy S = \(\left\{-5;3\right\}\)

8) \(\sqrt{4x^2-9}=\sqrt{-20x-18}\) ĐK: \(\left|x\right|\text{≥}\dfrac{3}{2}\) hoặc x ≤ \(\dfrac{-9}{10}\)

<=> 4x2 - 9 = -20x - 18

<=> 4x2 - 9 + 20x + 18 = 0

<=> 4x2 + 20x + 9 =0

<=> 4x2 + 2x + 18x + 9 =0

<=> 2x(2x + 1) + 9(2x + 1) = 0

<=> (2x + 9) . (2x + 1) = 0

=> \(\left[{}\begin{matrix}2x+9=0\\2x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=-9\\2x=-1\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=\dfrac{-9}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy S = \(\left\{\dfrac{-9}{2};\dfrac{-1}{2}\right\}\)

9) \(\sqrt{x-2}=\sqrt{x-2}\) ĐK: x ≥ 2

<=> x - 2 = x - 2

<=> x - x = 2 - 2

=> 0x = 0 với mọi x TMĐK: x ≥ 2

Kết luận: Phương trình vô nghiệm thoả mãn: x ≥ 2

1,

√(x-1) = √(2x+3)

->(√x-1)^2 = (√2x+3)^2

->x-1=2x+3

->x=-4

2

√(2x−3)=√(x−1) (bình phương lên tiếp)

->2x-3=x-1

->x=2

3->9 tự làm nha tương tự