Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
@Nguyễn Huy Thắng@Mysterious Person@bảo nam trần@Lightning Farron@Thiên Thảo@Sky SơnTùng
a) ĐKXĐ : \(x\ge5\)
Đặt \(\sqrt{x-5}=a;\sqrt[3]{3-x}=b\)(a \(\ge0\))
Khi đó phương trình thành a + b = 2
Lại có \(b^3+a^2=-2\)
=> HPT : \(\hept{\begin{cases}a+b=2\\b^3+a^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+\left(2-b\right)^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+b^2-4b+6=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=2-b\\\left(b+3\right)\left(b^2-2b+2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5\\b=-3\end{cases}}\)(tm)
a = 5 => x = 30 (tm)
Vậy x = 30 là nghiệm phương trình
d) Ta có \(\sqrt{25x^2-20x+4}+\sqrt{25x^2-40x+16}=0\)
<=> \(\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-4\right)^2}=2\)
<=> |5x - 2| + |5x - 4| = 2
Lại có |5x - 2| + |5x - 4| = |5x - 2| + |4 - 5x| \(\ge\left|5x-2+4-5x\right|=2\)
Dấu "=" xảy ra <=> \(\left(5x-2\right)\left(4-5x\right)\ge0\Leftrightarrow\frac{2}{5}\le x\le\frac{4}{5}\)
Vậy \(\frac{2}{5}\le x\le\frac{4}{5}\)là nghiệm phương trình
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
b)\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
Đk:\(-\sqrt{10}\le x\le\sqrt{10}\)
\(pt\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}=\left(x-4\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-\left(x-4\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+3=0\\10-x^2=x^2-8x+16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\-2x^2+8x-6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\-\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-3\) (thỏa)
c)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+3}=\sqrt{x^2-x+1}+\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+3}}+\sqrt{x+3}-\sqrt{x^2-x+1}-\sqrt{x+1}=0\)
Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x+1}=b;\sqrt{x+3}=c\left(a,b,c>0\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+c-a-b=0\)
\(\Leftrightarrow\dfrac{\left(a-c\right)\left(b-c\right)}{c}=0\)
\(\Leftrightarrow\left(a-c\right)\left(b-c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-c=0\\b-c=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=c\\b=c\end{matrix}\right.\)
*)Xét \(a=c\)\(\Rightarrow\sqrt{x^2-x+1}=\sqrt{x+3}\)
\(\Rightarrow x^2-x+1=x+3\Rightarrow x=\dfrac{2\pm\sqrt{12}}{2}\) (thỏa)
*)Xét \(b=c\)\(\Rightarrow\sqrt{x+1}=\sqrt{x+3}\)
\(\Rightarrow x+1=x+3\Rightarrow-2=0\) (loại)
\(a,\sqrt{3-x}+\sqrt{2-x}=1\)
\(\Rightarrow\sqrt{3+x}=1-\sqrt{2-x}\)
\(\Rightarrow3+x=1-2\sqrt{2-x}+2-x\)
\(\Rightarrow2x+2\sqrt{2-x}=0\)
\(\Rightarrow x+\sqrt{2-x}=0\)
\(\Rightarrow2-x=\left(-x\right)^2\)
\(\Rightarrow2-x=x^2\)
\(\Rightarrow2-x^2-x=0\)
\(\Rightarrow x^2+x-2=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
Vậy....
\(\sqrt{3x^2-5x+1}-\sqrt{3x^2-3x-3}\)=\(\sqrt{x^2-2}-\sqrt{x^2-3x+4}\) (dk tu xd)
\(\Leftrightarrow\frac{-2x+4}{\sqrt{3x^2-5x+1}+\sqrt{3x^2-3x-3}}\)=\(\frac{3x-6}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}+\frac{2}{\sqrt{3x^2-5x+1}+\sqrt{3x^2-3x-3}}\right)=0\)
\(\Leftrightarrow x=2\)
a) điều kiện 10 < hoặc bằng x < hoặc bằng 30
VT = căn (x-10) + căn (x-30) nhỏ hơn hoặc bằng căn (12+12 )*( x-10 +30-x) = 2 căn 10
VP = (x-20)2 + 2 căn 10
pt có nghiệm <=> x-10 = x-30 và x-20=0 <=> x = 20