Giải bất phương trình :
\(3\left(x^2-2\right)+\frac{4\sqrt{2}}{\sqrt{x^2-x+1}}>\sqrt{x}\left(\sqrt{x-1}+3\sqrt{x^2-1}\right)\)
Giải phương trình:
\(\frac{2\left(x-\sqrt{3}\right)\left(x-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
Giải phương trình sau :
\(\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]=\frac{2}{\sqrt{3}}+\sqrt{\frac{1-x^2}{3}}\)
ĐKXĐ: \(-1\le x\le1\)
Xét \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(1+x\right)+\left(1-x\right)+\sqrt{\left(1+x\right)\left(1-x\right)}\right]\)
\(=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)\)
Khi đó phương trình đề trở thành:
\(\sqrt{1+\sqrt{1-x}}\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)=\frac{2+\sqrt{1-x^2}}{3}\)
Vì \(2+\sqrt{1-x^2}>0\)nên ta có thể chia 2 vế cho \(2+\sqrt{1-x^2}\):
\(\Rightarrow\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{1+x}-\sqrt{1-x}\right)=\frac{1}{\sqrt{3}}\),Bình phương 2 vế:
\(\Rightarrow\left(1+\sqrt{1-x^2}\right)\left[\left(1+x\right)+\left(1-x\right)-2\sqrt{\left(1+x\right)\left(1-x\right)}\right]=\frac{1}{3}\)
\(\Leftrightarrow\left(1+\sqrt{1-x^2}\right)\left(2-2\sqrt{1-x^2}\right)=\frac{1}{3}\Leftrightarrow2\left(1+\sqrt{1-x^2}\right)\left(1-\sqrt{1-x^2}\right)=\frac{1}{3}\)\(\Leftrightarrow1-\left(1-x^2\right)=\frac{1}{3}\Leftrightarrow x^2=\frac{1}{6}\Leftrightarrow x=\pm\frac{1}{\sqrt{6}}\)
Ta xét phương trình đề: vế phải luôn không âm vì vậy vế trái phải không âm
Khi đó \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\ge0\Leftrightarrow1+x\ge1-x\Leftrightarrow x\ge0\)
Vậy ta chỉ nhận nghiệm duy nhất là \(x=\frac{1}{\sqrt{6}}\)
giải phương trình :
\(9\left(\sqrt{x+1}+\sqrt{x-2}\right)+1=4\left(\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-2\right)^3}\right)\)
Giải phương trình sau
a,\(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x+2\right)\sqrt{\frac{x-1}{x-2}}=3\)
b,\(\left(x-2\right)\left(x+2\right)+4\left(x-2\right)\sqrt{\frac{x+2}{x-2}}=-3\)
c, \(\frac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\frac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=-2\)
Sorry nha nhưng em mới học lớp 7 thôi à ~~
giải bất phương trình \(\left(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\right)\left(x^6-x^3+x^2-x+1\right)\ge0\)
Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)
Giải các phương trình sau:
a \(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
b \(\sqrt[3]{\left(65+x\right)^2}+4\sqrt[3]{\left(65-x\right)^2}=5\sqrt[3]{65^2-x^2}\)
c \(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
d \(\sqrt[3]{x-2}+\sqrt[3]{x+3}=\sqrt[3]{2x+1}\)
e \(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}\)
a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x+1}=b\end{matrix}\right.\)
\(\Rightarrow a+b=1+ab\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=1\\b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=1\\x+1=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
giải pt
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1
Giải phương trình :
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\) .
Dùng liên hợp.
pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)
\(-3\left(x-1\right)\left(x-\sqrt{3}\right)\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}\right)\)
\(+2\left(x-1\right)\left(x-\sqrt{2}\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+\sqrt{2}\right)=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)
\(=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(x+\sqrt{3}\right)\left(1-\sqrt{2}\right)\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left(x+1\right)\left(\sqrt{2}-\sqrt{3}\right)=3x-1\)
<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)
<=> \(x^2-3x+2=0\) phương trình bậc 2.
Em làm tiếp nhé!
Bài Toán :
Giải phương trình sau :
\(\frac{3\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\frac{4.\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{5\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)