gpt:
1, \(\sqrt{x^2+x+2}=\frac{x^2+5x+2}{2\left(x+1\right)}\)
2, \(\left(x+1\right)\sqrt{x^2-2x+3}=x^2+1\).
Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
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
gpt:
\(3\left(x^2-3x+1\right)+\sqrt{3\left(x^4+x^2+1\right)}=0\)
\(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)
Giải phương trình 1, \(x^2+9x+7=\left(2x+1\right)\sqrt{2x^2+4x+5}\)
2, GPT \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
3. GHPT \(\left\{{}\begin{matrix}x^2-2y-1=2\sqrt{5y+8}+\sqrt{7x-1}\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
ĐKXĐ: ...
Từ pt dưới:
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow y=x-2\)
Thế vào pt trên:
\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)
\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)
\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)
\(\Leftrightarrow x^2-5x+2=0\)
gpt a/ \(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)
b/ \(2\sqrt{1-x}-\sqrt{1+x}+3\sqrt{1-x^2}=3-x\)
b) Đặt \(u=\sqrt{1-x}\); \(v=\sqrt{1+x}\)
phương trình trở thành
\(2u-v+3uv=u^2+2\)\(\Rightarrow u^2-2u+v-3uv+2=0\)
lại có \(u^2+v^2=2\)
\(\Rightarrow u^2-2u-3uv+v+u^2+v^2=0\)
\(\Rightarrow\left(u-v-1\right)\left(2u-v\right)=0\)
đến đây thì easy rồi
a)
Đặt \(\sqrt{2x+1}=t\) ;\(\sqrt{x}=k\)
Phương trình trở thành
\(\left(3k^2+t^2\right)t-\left(3t^2+k^2\right)k-1=0\)
\(\Leftrightarrow3k^2t+t^3-3t^2k-k^3-1=0\)
\(\Leftrightarrow\left(t-k\right)\left(t^2+kt+k^2\right)-3tk\left(t-k\right)-1=0\)
\(\Leftrightarrow\left(t-k\right)^3-1=0\)
\(\Leftrightarrow\left(t-k-1\right)\left(\left(t-k\right)^2+t-k+1\right)=0\)
do t > k => t - k > 0
\(\Rightarrow\left(t-k\right)^2+t-k+1>0\)
\(\Rightarrow t-k-1=0\)
\(\Leftrightarrow t=1+k\)\(\Leftrightarrow\sqrt{2x+1}=1+\sqrt{x}\)
\(\Leftrightarrow2x+1=x+2\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
END
a)\(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)
ĐK:\(x\ge 0\)
\(\Leftrightarrow\left(5x+1\right)\sqrt{2x+1}-\left(\dfrac{31}{2}x+1\right)-\left(\left(7x+3\right)\sqrt{x}-\dfrac{31}{2}x\right)=0\)
\(\Leftrightarrow\dfrac{\left(5x+1\right)^2\left(2x+1\right)-\left(\dfrac{31}{2}x+1\right)^2}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{x\left(7x+3\right)^2-\left(\dfrac{31}{2}x\right)^2}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)
\(\Leftrightarrow\dfrac{\dfrac{1}{4}x\left(200x+19\right)\left(x-4\right)}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{\dfrac{1}{4}x\left(x-4\right)\left(196x-9\right)}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)
\(\Leftrightarrow\dfrac{1}{4}x\left(x-4\right)\left(\dfrac{200x+19}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{196x-9}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Nghe t đi phần nào khó cho qua :)) b tương tự
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
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 \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
GPT:
1, \(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)
2,\(x+3+\sqrt{1-x^2}=3\sqrt{x+1}+\sqrt{1-x}\)
ĐKXĐ:...
a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành:
\(3a^2-2b^2+ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow3a=2b\)
\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
Phương trình trở thành:
\(a^2+2+ab=3a+b\)
\(\Leftrightarrow a^2-3a+2+ab-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
1, gpt
a,\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
b, \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
c,\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
2/ cho x,y,z thỏa mãn : \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)
tính giá trị biểu thức B=\(\left(x^{29}+y^{29}\right)\left(x^{11}+y^{11}\right)\left(x^{2013}+y^{2013}\right)\)
ráng làm nốt rồi đi ngủ thoyy
1.
a) ĐK: \(x\ge2\)
\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)
Vậy...
b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)
Vậy...
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
Nhân cả 2 vế với \(\sqrt{2}\) ta được :
\(pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)
2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)
Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)
Tương tự 2 trường hợp còn lại ta đều được \(B=0\)
Vậy \(B=0\)
Gpt:
Bài quá dễ tự làm đi
k mình mình giải cho
Bạn nói dễ mà bạn không chịu làm thì bạn nói làm gì ???