\(\left(x^2-5x+6\right).\sqrt{1-x}=0\)

\(\Leftrightarrow\left(x^2-2x-3x+6\right).\sqrt{1-x}=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\x-3=0\\\sqrt{1-x}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\x=3\\x=1\end{cases}}\)

Vậy phương trình có tập nghiệm \(S=\left\{1;2;3\right\}\).

(x²-5x+6).\(\sqrt{1-x}=0\)

ĐK: 1-x >=0 <=> x=<1 (*)

(1) => \(\orbr{\begin{cases}x^2-5x+6=0\\\sqrt{1-x}=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(x-2\right)\left(x-3\right)\\1-x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2;x=3\\x=1\end{cases}}}\)

Các giá trị x=2; x=3 ktmđk (*)

Vậy x=1

giusp mình nhé mình đang cần lắm

a) \(\sqrt{8x^3}\cdot2x\)

\(=\sqrt{8x^3\cdot2x}\)

\(=\sqrt{16x^4}\)

\(=\sqrt{\left(4x^2\right)^2}\)

\(=4x^2\)

b) \(\sqrt{12x^5}\cdot\sqrt{3x}\)

\(=\sqrt{12x^5\cdot3x}\)

\(=\sqrt{36x^6}\)

\(=\sqrt{\left(6x^3\right)^2}\)

\(=\left|6x^3\right|\)

\(=6x^3\)

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

\(\left(\text{Đ}\text{KXĐ}:x\ge3\right)\)

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

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

\(\Leftrightarrow\Leftrightarrow\sqrt{x-2}+3\sqrt{x-3}-1=0\)

\(\Leftrightarrow\dfrac{x-2-1}{\sqrt{x-2}+1}+3\sqrt{x-3}=0\)

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

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

Pt \(\dfrac{1}{\sqrt{x-2}+1}+\dfrac{3}{\sqrt{x-3}}\) vô n_o

=> x - 3 = 0

<=> x = 3 (nhận)

sửa lẹ t làm cho

Ta có

\(1P=\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{x\sqrt{x}-1}{\sqrt{x}+1}-\sqrt{x}\right)\)

\(=\frac{1}{\sqrt{x}-1}.\frac{x\sqrt{X}-x-\sqrt{x}-1}{\sqrt{x}+1}\)

\(=1\frac{x\sqrt{x}-x-\sqrt{x}-1}{x-1}\)

Ta có thao câu b thì 1 - x > 0

<=> x < 1

=> \(0\le x< 1\)

Ta có \(P\sqrt{1-x}=\frac{x\sqrt{x}-x-\sqrt{x}-1}{-\sqrt{1-x}}< 0\)

\(\Leftrightarrow x\sqrt{x}-x-\sqrt{x}-1>0\)

Ta thấy \(0\le x< 1\Rightarrow x\sqrt{x}< x+\sqrt{x}+1\)

Vậy không có giá trị nào của x để cái trên xảy ra

Sửa đề: \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)

\(\Leftrightarrow\left|\sqrt{x-1}+2\right|+\left|\sqrt{x-1}-3\right|=5\)

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

TH1: x>=10

Pt sẽ là \(\sqrt{x-1}+\sqrt{x-1}-3=3\)

=>2 căn x-1=6

=>x-1=9

=>x=10

TH2: 1<=x<10

Pt sẽ là \(\sqrt{x-1}+3-\sqrt{x-1}=3\)

=>3=3(nhận)

a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)

b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)

\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)

c: \(=\sqrt{xy}+x\cdot\dfrac{\sqrt{y}}{\sqrt{x}}-y\cdot\dfrac{\sqrt{x}}{\sqrt{y}}\)

\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)