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PA
16 tháng 8 2017
\(\sqrt{x^2-3x+2}-\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\left(\sqrt{x^2-3x+2}-\sqrt{x-2}\right)-\left(\sqrt{x^2+2x-3}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)-\left(x-2\right)}{\sqrt{x^2-3x+2}+\sqrt{x-2}}-\dfrac{\left(x^2+2x-3\right)-\left(x+3\right)}{\sqrt{x^2+2x-3}-\sqrt{x+3}}=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x-2}}-\dfrac{\left(x-2\right)\left(x+3\right)}{\sqrt{\left(x+3\right)\left(x-1\right)}-\sqrt{x+3}}=0\)
\(\Leftrightarrow\left(x-2\right)\left[\dfrac{x-2}{\sqrt{x-2}\left(\sqrt{x-1}+1\right)}-\dfrac{x+3}{\sqrt{x+3}\left(\sqrt{x-1}-1\right)}\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\right]=0\)
Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô no
(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))
=> x - 2 = 0
<=> x = 2 (nhận)
PA
16 tháng 8 2017
\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)
\(\Leftrightarrow\dfrac{\left(4x+1\right)-\left(3x-2\right)}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}\right)\left(x+3\right)=0\)
TH1:
x + 3 = 0
<=> x = - 3 (loại)
TH2:
\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)
\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\)
\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)+\left(\sqrt{3x-2}-2\right)=0\)
\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)
\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{3\left(x-2\right)}{\sqrt{3x-2}+2}=0\)
\(\Leftrightarrow\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}\right)\left(x-2\right)=0\)
Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô no
=> x - 2 = 0
<=> x = 2 (nhận)
~ ~ ~
Vậy x = 2
3 tháng 9 2019
Trả lời :
Con a giai pt vế trái rồi nhân căn bình phương cả 2 vế
Con b cũng giải pt vế phải chuyển vế rồi bình phương cả 2 vế
Chắc vậy
k bt
VT
2
21 tháng 7 2019
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
1 tháng 7 2019
tth, Hoàng Tử Hà, Bonking, Quoc Tran Anh Le, Vũ Huy Hoàng,
Akai Haruma, @Nguyễn Việt Lâm
giúp mk vs! ngày mai phải nộp r
ĐKXĐ: x>=\(-\frac{1}{3}\)
PT \(\Leftrightarrow x^2+2x-3=\sqrt{2x+2}-\sqrt{3x+1} \)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=\frac{2x+2-3x-1}{\sqrt{2x+2}+\sqrt{3x+1}}\Leftrightarrow\left(x-1\right)\left(x+3+\frac{1}{\sqrt{2x+2}+\sqrt{3x+1}}\right)=0\)
dễ nhận thấy x=1 là nghiệm duy nhất của pt