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giải các phương trình a)\(\sqrt{2x-1}+x^2-3x+1=0\)
b)\(\sqrt{x-1}+\sqrt{9-x}+2\sqrt{-x^2+10x-9}=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(ĐK:x\ge\frac{1}{2}\)
PT\(\Leftrightarrow-\left(2x-1-\sqrt{2x-1}+\frac{1}{4}\right)+x^2-x+\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\left(\sqrt{2x-1}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\sqrt{2x-1}-1\right)\left(x-\sqrt{2x-1}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\sqrt{2x-1}=1\\x=\sqrt{2x-1}\end{cases}}\)
Chắt duoc roi he
b.
\(ĐK:1\le x\le9\)
\(\Rightarrow\hept{\begin{cases}a+b+2ab=12\\a^2+b^2=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2+\left(a+b\right)-12=0\\a^2+b^2=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b-3\right)\left(a+b+4\right)=0\\a^2+b^2=7\end{cases}}\)
Loai \(a+b+4=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vừa làm bên Học 24 xong nhưng do gửi link thì bị lỗi nên t up lại, tiện thể ăn điểm luôn (tất nhiên giúp you vẫn là lí do chính, điểm là tiện thôi :))
\(pt\left(2\right)\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+12}-\sqrt{y}-\sqrt{x+2}=0\)
\(\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+12}-2\sqrt{y}-\left(\sqrt{x+2}-\sqrt{y}\right)=0\)
\(\Leftrightarrow\frac{2\left(x-y\right)^2+10x-6y+12-4y}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\frac{x+2-y}{\sqrt{x+2}+\sqrt{y}}=0\)
\(\Leftrightarrow\frac{2\left(x-y+3\right)\left(x-y+2\right)}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\frac{x-y+2}{\sqrt{x+2}+\sqrt{y}}=0\)
\(\Leftrightarrow\left(x-y+2\right)\left(\frac{2\left(x-y+3\right)}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\frac{1}{\sqrt{x+2}+\sqrt{y}}\right)=0\)
\(\Rightarrow x=y-2\). Thay vào \(pt\left(1\right)\) ta có:
\(pt\left(1\right)\Leftrightarrow\sqrt{y^2-8\left(y-2\right)+9}-\sqrt[3]{\left(y-2\right)y+12-6\left(y-2\right)}\le1\)
\(\Leftrightarrow\sqrt{y^2-8y+25}-\sqrt[3]{y^2-8y+24}\le1\)
\(\Leftrightarrow\left(\sqrt{y^2-8y+25}-3\right)-\left(\sqrt[3]{y^2-8y+24}-2\right)\le0\)
\(\Leftrightarrow\frac{y^2-8y+25-9}{\sqrt{y^2-8y+25}+3}-\frac{y^2-8y+24-8}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\le0\)
\(\Leftrightarrow\frac{\left(y-4\right)^2}{\sqrt{y^2-8y+25}+3}-\frac{\left(y-4\right)^2}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\le0\)
\(\Leftrightarrow\left(y-4\right)^2\left(\frac{1}{\sqrt{y^2-8y+25}+3}-\frac{1}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\right)\le0\)
\(\Rightarrow y=4\Rightarrow x=y-2=4-2=2\)
Vậy \(x=2;y=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(\sqrt{\left(x+3\right)^2}\)+ \(\sqrt{\left(x+4\right)^2}\)+\(\sqrt{\left(x+5\right)^2}\)=9x
=> x + 3 + x + 4 + x + 5 = 9x
=> - 6x = - 12
=> x=2
Ủa sao phá đc trị tuyệt đối hay v bạn? (căn a^2 = trị tuyệt đối của a )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{25x^2-10x+1}=4x+9\)
\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x+9\)
\(\Leftrightarrow\left|5x-1\right|=4x+9\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=4x+9\\5x-1=-4x-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-\frac{8}{9}\end{cases}}}\)
Vậy ...
\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2}=\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2}-\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}.\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+1}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
b, ĐK \(x\ge-4\)
PT
<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)
<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)
Giải (2)
=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)
<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)
<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)
<=> \(x^2-7x-4=6\sqrt{x+4}\)
<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)
Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)
=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)
=> \(a^2-b^2+6\left(a-b\right)=0\)
<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)
+ a=b
=> \(x-6=\sqrt{x+4}\)
=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)
+ a+b+6=0
=> \(x+\sqrt{x+4}=0\)(loại)
Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
À câu a mình tự làm được rồi nhé! Các bạn chỉ cần làm câu b cho mình là được.
b, \(\frac{2\sqrt{x}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\)
ĐK \(x\ge0\)
Pt
<=> \(2\sqrt{x}+\sqrt{x\left(x+1\right)}=\sqrt{\left(x+1\right)\left(x+9\right)}\)
<=> \(4x+x^2+x+4\sqrt{x^2\left(x+1\right)}=x^2+10x+9\)
<=> \(4x\sqrt{x+1}=5x+9\)
<=> \(16x^2\left(x+1\right)=25x^2+90x+81\)với mọi \(x\ge0\)
<=> \(16x^3-9x^2-90x-81=0\)
<=> \(x=3\)(tm ĐK)
Vậy x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}=8\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\)
\(\Rightarrow x-3+x+5=8\)
\(\Rightarrow2x=6\Rightarrow x=3\)
\(\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}=8\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\Leftrightarrow\left|x-3\right|+\left|x+5\right|=8\) (1)
Nếu \(x< -5\) thì (1) trở thành:
\(3-x+\left(-x-5\right)=8\Leftrightarrow-2x-2=8\Leftrightarrow x=-5\) (loại)
-Nếu \(-5\le x< 3\) thì (1) trở thành:
\(3-x+x+5=8\Leftrightarrow8=8\)
-Nếu \(x>3\) thì (1) trở thành:
\(x-3+x+5=8\Leftrightarrow2x+2=8\Leftrightarrow x=3\) (thỏa mãn)
Vậy \(-5\le x\le3\)
ĐKXĐ : \(1\le x\le9\)
\(\sqrt{x-1}+\sqrt{9-x}+2\sqrt{-x^2+10x-9}=12\)
<=> \(\sqrt{x-1}+\sqrt{9-x}+2\sqrt{\left(x-1\right)\left(9-x\right)}=12\)(*)
Đặt t = \(\sqrt{x-1}+\sqrt{9-x}\) ( t\(\ge\)0 ) => t2 = \(x-1+9-x+2\sqrt{\left(x-1\right)\left(9-x\right)}\)
<=> t2 = \(2\sqrt{\left(x-1\right)\left(9-x\right)}+8\)
<=> \(2\sqrt{\left(x-1\right)\left(9-x\right)}=t^2-8\) . Thay vào pt (*), ta có :
t + t2 - 8 = 12
<=> t2 + t - 20 = 0
<=> ( t + 1/2 )2 = 81/4
<=>\(\orbr{\begin{cases}t+\frac{1}{2}=\frac{9}{2}\\t+\frac{1}{2}=-\frac{9}{2}\end{cases}}\) <=> \(\orbr{\begin{cases}t=4\left(tm\right)\\t=-5\left(ktm\right)\end{cases}}\)
Suy ra : \(2\sqrt{\left(x-1\right)\left(9-x\right)}=4^2-8=8\)
<=> \(\sqrt{\left(x-1\right)\left(9-x\right)}=4\)
<=> ( x - 1 ) ( 9 - x ) = 16
<=> - x2 + 10x - 25 = 0
<=> - ( x - 5 )2 = 0
<=> x = 5 ( tmdk )
Vậy pt có tập nghiệm S = { 5 }