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nhìn mà nhác giải vl :v
a) \(\sqrt{3x^2-2x+1}+4x=\sqrt{3x^2+2x}+1\)
<=> \(\sqrt{3x^2-2x+1}=\sqrt{3x^2+2x}+1-4x\)
<=> \(\left(\sqrt{3x^2-2x+1}\right)^2=\left(\sqrt{3x^2+2x}+1-4x\right)^2\)
<=> \(3x^2-2x+1=19x^2-8\sqrt{3x^2+2x}.x-6x+2\sqrt{3x^2+2x}+1\)
<=> \(-16x^2+8\sqrt{3x^2+2x}.x+4x-2\sqrt{3x^2+2x}=0\)
<=> \(-2\left(4x-1\right)\left(2x-\sqrt{3x^2+2x}\right)=0\)
<=> \(\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=2\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=0\end{cases}}\) (vì k có ngoặc vuông 3 nên mình dùng tạm ngoặc nhọn, thông cảm)
<=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=2\end{cases}}\)
b) \(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)
<=> \(\sqrt{x^2+x-2}=\sqrt{2\left(x-1\right)}+1-x^2\)
<=> \(\left(\sqrt{x^2+x-2}\right)^2=\left[\sqrt{2\left(x-1\right)}+1-x^2\right]^2\)
<=> \(x^2+x-2=x^4-2\sqrt{2}.x^2.\sqrt{x-1}-2x^2+2x+2\sqrt{2}.\sqrt{x-2}-1\)
<=> \(x^4-2\sqrt{2}.x^2.\sqrt{x-1}-2x^2+2x+2\sqrt{2}.\sqrt{x-1}-1=x^2+x-2\)
<=> \(-2\sqrt{2}.x^2.\sqrt{x-1}+2\sqrt{2}.\sqrt{x-1}-1=-x^4+3x^2-x-2\)
<=> \(-2\sqrt{2}.x^2.\sqrt{x-1}+2\sqrt{2}.\sqrt{x-1}=-x^4+3x^2-x-1\)
<=> \(-2\sqrt{2}.\sqrt{x-1}.\left(x^2+1\right)=-x^4+3x^2-x-1\)
<=> \(\left[-2\sqrt{2}.\sqrt{x-1}\left(x^2+1\right)\right]^2=\left(-x^4+3x^2-x-1\right)^2\)
<=> \(8x^5-8x^4-16x^3+16x^2+8x-8=x^8-6x^6+2x^5+11x^4-6x^3-5x^2+2x+1\)
<=> x = 1
d) mình làm tắt cho nhanh
d) \(\left(\sqrt{4+x}-1\right)\left(\sqrt{1-x}+1\right)=2x\)
<=> \(\sqrt{4+x}.\sqrt{x-1}+\sqrt{4+x}-\sqrt{x-1}-1=2x\)
<=> \(\sqrt{4+x}.\sqrt{1-x}+\sqrt{4+x}-\sqrt{1-x}=2x+1\)
<=> \(\sqrt{4+x}.\sqrt{x-1}+\sqrt{4+x}=2x+1+\sqrt{x-1}\)
<=> \(\left(\sqrt{4+x}.\sqrt{1-x}+\sqrt{4+x}\right)^2=\left(2x+1+\sqrt{1-x}\right)^2\)
<=> \(2\sqrt{-x+1}.\left(x+4\right)=5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6\)
<=> \(\frac{2\sqrt{-x+1}.\left(x+4\right)}{2\left(x+4\right)}=\frac{5x^2}{2\left(x+4\right)}+\frac{4x\sqrt{-x+1}}{2\left(x+4\right)}+\frac{5x}{2\left(x+4\right)}+\frac{2\sqrt{-2x+1}}{2\left(x+4\right)}-\frac{6}{2\left(x+4\right)}\)
<=> \(\sqrt{-x+1}=\frac{5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6}{2\left(4+x\right)}\)
<=> \(2\sqrt{-x+1}.\left(4+x\right)=5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6\)
<=> \(-2x\sqrt{-x+1}+6\sqrt{-x+1}=5x^2+5x-6\)
<=> \(\frac{2\sqrt{-x+1}.\left(-x+3\right)}{2\left(-x+3\right)}=\frac{5x^2}{2\left(-x+3\right)}+\frac{5x}{2\left(-x+3\right)}-\frac{6}{2\left(-x+3\right)}\)
<=> \(\sqrt{-x+1}=\frac{5x^2+5x-6}{2\left(x-3\right)}\)
<=> \(\left(\sqrt{-x+1}\right)^2=\left[\frac{5x^2+5x-6}{2\left(3-x\right)}\right]^2\)
<=> \(-x+1=\frac{25x^4+50x^3-35x^2-60x+36}{36-24+4x}\)
<=> \(\hept{\begin{cases}x=0\\x=\frac{21}{25}\\x=-3\end{cases}}\)=> x = 21/25 (lý do dùng ngoặc nhọn như lý do mình ghi ở trên =))) )
=> x = 21/25
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a) ĐK: \(x\geq \frac{1}{2}\)
Ta có: \(\sqrt{2x-1}-\sqrt{x+1}=2x-4\)
\(\Leftrightarrow \frac{(2x-1)-(x+1)}{\sqrt{2x-1}+\sqrt{x+1}}=2(x-2)\)
\(\Leftrightarrow \frac{x-2}{\sqrt{2x-1}+\sqrt{x+1}}=2(x-2)\)
\(\Leftrightarrow (x-2)\left(\frac{1}{\sqrt{2x-1}+\sqrt{x+1}}-2\right)=0\)
\(\Rightarrow \left[\begin{matrix} x-2=0\leftrightarrow x=2\\ \frac{1}{\sqrt{2x-1}+\sqrt{x+1}}=2(*)\end{matrix}\right.\)
Đối với $(*)$:
Vì \(x\geq \frac{1}{2}\Rightarrow \sqrt{2x-1}+\sqrt{x+1}\geq \sqrt{\frac{1}{2}+1}>1\)
\(\Rightarrow \frac{1}{\sqrt{2x-1}+\sqrt{x+1}}< 1\)
Do đó $(*)$ vô nghiệm
Vậy pt có nghiệm duy nhất $x=2$
b) ĐK:.....
\(\sqrt{2x^2-3x+10}+\sqrt{2x^2-5x+4}=x+3\)
TH1:
\(\sqrt{2x^2-3x+10}=\sqrt{2x^2-5x+4}\)
\(\Rightarrow 2x^2-3x+10=2x^2-5x+4\)
\(\Rightarrow 2x+6=0\Rightarrow x=-3\) (thử lại thấy không thỏa mãn)
TH2: \(\sqrt{2x^2-3x+10}\neq \sqrt{2x^2-5x+4}\), tức là \(x\neq -3\)
PT ban đầu tương đương với:
\(\frac{(2x^2-3x+10)-(2x^2-5x+4)}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=x+3\)
\(\Leftrightarrow \frac{2(x+3)}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=x+3\)
\(\Leftrightarrow \frac{2}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=1\) (do \(x\neq -3\) )
\(\Rightarrow \sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}=2\)
\(\Rightarrow \sqrt{2x^2-3x+10}=2+\sqrt{2x^2-5x+4}\)
Bình phương 2 vế:
\(2x^2-3x+10=4+2x^2-5x+4+4\sqrt{2x^2-5x+4}\)
\(\Leftrightarrow x+1=2\sqrt{2x^2-5x+4}\)
\(\Rightarrow (x+1)^2=4(2x^2-5x+4)\)
\(\Rightarrow 7x^2-22x+15=0\Rightarrow \left[\begin{matrix} x=\frac{15}{7}\\ x=1\end{matrix}\right.\) (thử đều thấy t/m)
Vậy...........
a) Ta có: \(\sqrt{3x-2}=x+1\) ( ĐK: \(x\ge\frac{2}{3}\))
\(\Leftrightarrow\left(\sqrt{3x-2}\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow3x-2=x^2+2x+1\)
\(\Leftrightarrow x^2-x+3=0\)
\(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\forall x\)
mà \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)
\(\Rightarrow\)\(S=\varnothing\)
b) Ta có: \(\sqrt{x^2-2x+1}=x-1\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=x-1\)
\(\Leftrightarrow\left|x-1\right|=x-1\)
+ Với \(x< 1\)\(\Rightarrow\)\(\left|x-1\right|=1-x\)
Ta có: \(1-x=x-1\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\left(L\right)\)
+ Với \(x\ge1\)\(\Rightarrow\)\(\left|x-1\right|=x-1\)
Ta có: \(x-1=x-1\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow\)\(x\inℝ\)
Vậy \(S=ℝ\)
c) Ta có: \(\sqrt{2x+1}=x-2\) ( ĐK: \(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(\sqrt{2x+1}\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow2x+1=x^2-4x+4\)
\(\Leftrightarrow x^2-6x+3=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)-6=0\)
\(\Leftrightarrow\left(x-3\right)^2=6\)
\(\Leftrightarrow x-3=\pm\sqrt{6}\)
+ \(x-3=\sqrt{6}\)\(\Leftrightarrow\)\(x=3+\sqrt{6}\approx5,45\)\(\left(TM\right)\)
+ \(x-3=-\sqrt{6}\)\(\Leftrightarrow\)\(x=3-\sqrt{6}\approx0,55\)\(\left(TM\right)\)
Vậy \(S=\left\{5,45;0,55\right\}\)
d) Ta có: \(\sqrt{x^2-3}=x^2-3\) ( ĐK: \(x\ge\pm\sqrt{3}\))
\(\Leftrightarrow\sqrt{x^2-3}-\left(\sqrt{x^2-3}\right)^2=0\)
\(\Leftrightarrow\sqrt{x^2-3}.\left(1-\sqrt{x^2-3}\right)=0\)
+ \(\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(x^2-3=0\)\(\Leftrightarrow\)\(x^2=3\)\(\Leftrightarrow\)\(x=\pm\sqrt{3}\)\(\left(TM\right)\)
+ \(1-\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(\sqrt{x^2-3}=1\)
\(\Leftrightarrow\)\(x^2-3=1\)
\(\Leftrightarrow\)\(x^2=4\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\left(TM\right)\\x=-2\left(L\right)\end{cases}}\)
Vậy \(S=\left\{-\sqrt{3};\sqrt{3};2\right\}\)
e) Ta có: \(\sqrt{x^2-6x+9}=6-x\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)
\(\Leftrightarrow\left|x-3\right|=6-x\)
+ Với \(x< 3\)\(\Leftrightarrow\)\(\left|x-3\right|=3-x\)
Ta có: \(3-x=6-x\)
\(\Leftrightarrow0x=3\)( vô nghiệm )
+ Với \(x\ge3\)\(\Leftrightarrow\)\(\left|x-3\right|=x-3\)
Ta có: \(x-3=6-x\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{9}{2}\)\(\left(TM\right)\)
Vậy \(S=\left\{\frac{9}{2}\right\}\)
g) Ta có: \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{x-1}-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}-1\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
+ Với \(x< 2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1< 0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\)
Ta có: \(1-\sqrt{x-1}=\sqrt{x-1}-1\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)\(\left(L\right)\)
+ Với \(x\ge2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1\ge0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Ta có: \(\sqrt{x-1}-1=\sqrt{x-1}-1\)
\(\Leftrightarrow0x=0\)( vô số nghiệm )
Vậy \(S=ℝ\)