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a) chắc là nhóm lại thui để sau mk làm:v
b)\(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
Đk: tự lm nhé :v
\(pt\Leftrightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}-\left(\sqrt{2x-1}-\sqrt{3}\right)=2x^2-8\)
\(\Leftrightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2x-1-3}{\sqrt{2x-1}+\sqrt{3}}=2\left(x^2-4\right)\)
\(\Leftrightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}=2\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow\frac{\frac{-2\left(x-2\right)}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}-2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)\right)=0\)
Dễ thấy: \(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)< 0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\left(1\right)\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)
\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)
Chúc bạn học tốt !!!
\(x=\sqrt[3]{7+\sqrt{\frac{49}{8}}}+\sqrt[3]{7-\sqrt{\frac{49}{8}}}\)
ta lập phương hai vế có
\(x^3=7+\sqrt{\frac{49}{8}}+7-\sqrt{\frac{49}{8}}+3\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}x\)
\(< =>x^3=14+3\sqrt[3]{7^2-\frac{49}{8}}x\)
\(< =>x^3=14+3\sqrt[3]{\frac{343}{8}}x\)
\(< =>x^3=14+3.\frac{7}{2}x\)
\(< =>2x^3-21x-28=0\)
nên
\(fx=\left(2x^3-21x-29\right)^3=\left(2x^3-21x-28-1\right)^3=\left(-1\right)^3=-1\)
\(\Leftrightarrow\sqrt{4-\left(1-x\right)^2}=\sqrt{3}\)
\(\Leftrightarrow4-\left(1-x\right)^2=3\)
\(\Leftrightarrow4-\left(1-2x+x^2\right)-3=0\)
\(\Leftrightarrow4-1+2x-x^2-3=0\)
\(\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
vay x=0 ; x=2
\(\sqrt{3x^2-5=2}\left(x\ge\sqrt{\frac{5}{3}}\right)\)
\(\Leftrightarrow3x^2-5=4\)
\(\Leftrightarrow3x^2=9\Leftrightarrow x^2=3\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}\left(tm\right)\\x=-\sqrt{3}\left(kotm\right)\end{cases}}\)
vay \(x=\sqrt{3}\)
\(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\left(x\ge49\right)\)
\(\Leftrightarrow\sqrt{x-49}=2\Leftrightarrow x^2-98x+2401=4\)
\(\Leftrightarrow x^2-98x+2397=0\Leftrightarrow x^2-47x-51x+2397\)\(\Leftrightarrow x\left(x-47\right)-51\left(x-47\right)\Leftrightarrow\left(x-47\right)\left(x-51\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-51=0\\x-47=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=51\left(tm\right)\\x=47\left(kotm\right)\end{cases}}}\)
xay x=51
\(\sqrt{\frac{-6}{1+x}}=5\left(x< -1\right)\)
\(\Leftrightarrow\frac{36}{x^2+2x+1}=25\Leftrightarrow25x^2+50x+25=36\)
\(\Leftrightarrow25x^2+50x-11=0\Leftrightarrow25x^2-5x+55x-11\)
\(\Leftrightarrow5x\left(5x-1\right)+11\left(5x-1\right)\Leftrightarrow\left(5x-1\right)\left(5x+11\right)\)\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\5x+11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\left(kotm\right)\\x=\frac{-11}{5}\left(tm\right)\end{cases}}}\)
vay \(x=\frac{-11}{5}\)
nhung cau nay binh phuong len la xong
y 3 xem lai de bai
y 4,7 ko biet lam
Bài 3 nhé bạn đặt cái căn đầu là a ,căn sau là b
a+b=x
ab=1
Rồi tính lần lượt a3 +b3 bằng ẩn x hết
và mũ 4 cũng vậy rồi lấy 2 số nhân nhau .Bđ là ra
Em thử nha,sai thì thôi ạ.
2/ ĐK: \(-2\le x\le2\)
PT \(\Leftrightarrow\sqrt{2x+4}-\sqrt{8-4x}=\frac{6x-4}{\sqrt{x^2+4}}\)
Nhân liên hợp zô: với chú ý rằng \(\sqrt{2x+4}+\sqrt{8-4x}>0\) với mọi x thỏa mãn đk
PT \(\Leftrightarrow\frac{6x-4}{\sqrt{2x+4}+\sqrt{8-4x}}-\frac{6x-4}{\sqrt{x^2+4}}=0\)
\(\Leftrightarrow\left(6x-4\right)\left(\frac{1}{\sqrt{2x+4}+\sqrt{8-4x}}-\frac{1}{\sqrt{x^2+4}}\right)=0\)
Tới đây thì em chịu chỗ xử lí cái ngoặc to rồi..
1.\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)
ĐK \(x\ge-1\)
Nhân liên hợp ta có
\(\left(x+3-x-1\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
<=>\(x^2+\sqrt{\left(x+1\right)\left(x+3\right)}=x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
<=> \(\left(x^2-x\sqrt{x+3}\right)+\left(\sqrt{\left(x+1\right)\left(x+3\right)}-x\sqrt{x+1}\right)=0\)
<=> \(\left(x-\sqrt{x+3}\right)\left(x-\sqrt{x+1}\right)=0\)
<=> \(\orbr{\begin{cases}x=\sqrt{x+3}\\x=\sqrt{x+1}\end{cases}}\)
=> \(x\in\left\{\frac{1+\sqrt{13}}{2};\frac{1+\sqrt{5}}{2}\right\}\)
Vậy \(x\in\left\{\frac{1+\sqrt{13}}{2};\frac{1+\sqrt{5}}{2}\right\}\)
Tìm nhẩm nghiệm rồi nhân liên hợp