ta có:

\(\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow\left(16-2x+x^2-9+2x-x^2\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow7=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

Ta có:

\(\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)=7\)

\(\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow\left(16-2x+x^2-9+2x-x^2\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow7=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

Ủng hộ nha

cau tra loi duoc olm lua chon

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

\(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

\(\Leftrightarrow\dfrac{\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\)

\(\Leftrightarrow\dfrac{16-2x+x^2-9+2x-x^2}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\)

\(\Leftrightarrow\dfrac{7}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\Leftrightarrow\dfrac{7}{A}=1\Rightarrow A=7\)

Có: \(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2+15}-\sqrt{\left(x-1\right)^2+8}=1\)

\(\Leftrightarrow2\left(x-1\right)^2+23-2\sqrt{\left(x-1\right)^4+23\left(x-1\right)^2+120}=1\)

Đặt \(t=\left(x-1\right)^2\left(t\ge0\right)\)

\(\Rightarrow2t+23-2\sqrt{t^2+23t+120}=1\)

\(\Leftrightarrow t+11=\sqrt{t^2+23t+120}\)

\(\Leftrightarrow t^2+22t+121=t^2+23t+120\)

\(\Leftrightarrow t=1\left(TM\right)\)

\(\Rightarrow x\in\left\{0;2\right\}\)

Thay x=0 vào A, ta có:

\(A=\sqrt{16-2.0+0^2}+\sqrt{9-2.0+0^2}=7\)

Thay x=2 vào A, ta có:

\(A=\sqrt{16-2.1+1^2}+\sqrt{9-2.1+1^2}=\sqrt{15}+2\sqrt{2}\)

Ta có \(\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)=16-2x+x^2-\left(9-2x+x^2\right)=16-2x+x^2-9+2x-x=7\Leftrightarrow\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)=7\Leftrightarrow1.A=7\Leftrightarrow A=7\)

a, ĐKXĐ: \(x\le2\)

\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)

b, ĐKXĐ: \(x\ge-1\)

\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)

c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow4-2x=25\)

hay \(x=-\dfrac{21}{2}\)

c: \(\Leftrightarrow\left|2x+3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)

\(\Leftrightarrow x=6\)

\(a,ĐK:x\ge\dfrac{5}{2}\\ PT\Leftrightarrow2x-5=4\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\le4\\ PT\Leftrightarrow\left|x-8\right|=4-x\\ \Leftrightarrow\left[{}\begin{matrix}x-8=4-x\left(x\ge8\right)\\8-x=4-x\left(x\le8\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\left(trái.vs.ĐK\right)\\0x=4\left(ktm\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

a) \(\sqrt{2x-5}=2\)

\(\Leftrightarrow\) \(\sqrt{2x-5}^2=2^2\)

\(\Leftrightarrow\) \(2x-5=4\)

\(\Leftrightarrow\) 2x = 9

\(\Leftrightarrow\) x = \(\dfrac{9}{2}\)

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