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từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
a, ĐKXĐ :\(\left\{{}\begin{matrix}x\ge-3\\x\le6\end{matrix}\right.\)
=> \(-3\le x\le6\)
Ta có : \(\sqrt{x+3}+\sqrt{6-x}-\sqrt{\left(x+3\right)\left(6-x\right)}=3\)
- Đặt \(\sqrt{x+3}=a,\sqrt{6-x}=b\left(a,b\ge0\right)\)
=> \(a^2+b^2=x+3+6-x=9\)
Ta có : \(a+b-ab=3\)
=> \(a+b=ab+3\)
Ta có : \(\left(a+b\right)^2-2ab=9\)
=> \(\left(ab+3\right)^2-2ab=9\)
=> \(a^2b^2+6ab+9-2ab=9\)
=> \(a^2b^2+4ab=0\)
=> \(ab\left(ab+4\right)=0\)
=> \(\left[{}\begin{matrix}ab=0\\ab=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a+b=3\\a+b=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}a=-b+3\\a=-b-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b\left(-b+3\right)=0\\b\left(-b-1\right)=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b\left(-b+3\right)=0\\b^2+b-4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b=0\left(TM\right)\\b=3\left(TM\right)\\b=\frac{-1-\sqrt{17}}{2}\left(KTM\right)\\b=\frac{-1+\sqrt{17}}{2}\left(TM\right)\end{matrix}\right.\) => \(\left[{}\begin{matrix}6-x=0\\6-x=9\\6-x=\frac{9-\sqrt{17}}{2}\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=6\\x=-3\\x=\frac{3+\sqrt{17}}{2}\end{matrix}\right.\) ( TM )
=> \(\left[{}\begin{matrix}a=3\\a=0\\a=\frac{7-\sqrt{17}}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x+3=0\\x+3=9\\x+3=\frac{33-7\sqrt{17}}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=6\\x=\frac{27-7\sqrt{17}}{2}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)( TM )
Vậy ...
b, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ge0\\x+3\ge0\\4-2x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge1\\x\ge-3\\x\le2\end{matrix}\right.\)
=> \(1\le x\le2\)
Ta có : \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x+3\right)}=4-2x\)
- Đặt \(\sqrt{x-1}=a,\sqrt{x+3}=b\left(a,b\ge0\right)\)
=> \(a^2+b^2=\left(a+b\right)^2-2ab=x-1+x+3=2x+2\)
Ta được: \(a+b+2ab=4-2x=-2x-2+6=-\left(2x+2\right)+6=-a^2-b^2+6\)
=> \(a^2+2ab+b^2=-a-b=\left(a+b\right)^2=-\left(a+b\right)\)
=> \(\left(a+b\right)^2+\left(a+b\right)=0\)
=> \(\left(a+b\right)\left(a+b+1\right)=0\)
=> \(\left[{}\begin{matrix}a+b=0\\a+b+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x-1}+\sqrt{x+3}=0\\\sqrt{x+1}+\sqrt{x+3}=-1\left(VL\right)\end{matrix}\right.\)
Ta thấy : \(\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{x+3}\ge0\end{matrix}\right.\)
=> \(\sqrt{x-1}+\sqrt{x+3}\ge0\forall x\)
- Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\left(TM\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
=> x = 1 .
Vậy ...
\(\sqrt{28-6\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=3\sqrt{3}-1\)
\(\sqrt{6-\sqrt{20}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-1\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\left(x+2\right)}}\)
\(=\sqrt{\left(\sqrt{x+2}+\sqrt{x+1}\right)^2}\)
\(=\sqrt{x+2}+\sqrt{x+1}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{\left(x-1\right)\left(x+3\right)}+\left(x+3\right)}\)
\(=\sqrt{\left(\sqrt{x+3}-\sqrt{x-1}\right)^2}\)
\(=\left|\sqrt{x+3}-\sqrt{x-1}\right|\)
\(\sqrt{21-6\sqrt{6}}+\sqrt{21+6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\)
\(=6\sqrt{2}\)
\(M=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)\(\left[\dfrac{\left(\sqrt{x}+1\right)-\left(3-\sqrt{x}\right)}{\sqrt{x}+1}\right]\)
\(=\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right]\times\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}\times2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
\(\sqrt{3+x}+\sqrt{6-x}-\sqrt{\left(3+x\right)\left(6-x\right)}=3\)(ĐK: \(-3\le x\le6\))
Đặt \(t=\sqrt{3+x}+\sqrt{6-x}\ge0\)
\(\Leftrightarrow t^2=3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}\)
\(\Leftrightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=\frac{t^2-9}{2}\)
Phương trình ban đầu tương đương với:
\(t-\frac{t^2-9}{2}=3\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\left(tm\right)\\t=-1\left(l\right)\end{cases}}\)
Với \(t=3\):
\(\sqrt{3+x}+\sqrt{6-x}=3\)
\(\Leftrightarrow9+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\left(3+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\left(tm\right)\\x=6\left(tm\right)\end{cases}}\)
\(\sqrt{2x+1}-\sqrt{2-x}=x+3\)(ĐK: \(-\frac{1}{2}\le x\le2\))
Ta có: \(\sqrt{2x+1}-\sqrt{2-x}\le\sqrt{2x+1}\le\sqrt{2.2+1}=\sqrt{5}\)(vì \(-\frac{1}{2}\le x\le2\))
\(x+3\ge-\frac{1}{2}+3=2,5\)(vì \(-\frac{1}{2}\le x\le2\))
mà \(\sqrt{5}< 2,5\)
do đó phương trình vô nghiệm.