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PT 2
\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))
\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Rightarrow2x^2-3x+6=0\)
=> PT vô nghiệm.
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
giải phương trình sau
\(\dfrac{\text{x+1}}{11}-\dfrac{2x-5}{15}=\dfrac{3x-47}{17}-\dfrac{4x-59}{19}\)
Lời giải:
PT \(\Leftrightarrow (\frac{x+1}{11}-1)-(\frac{2x-5}{15}-1)=(\frac{3x-47}{17}+1)-(\frac{4x-59}{19}+1)\)
\(\Leftrightarrow \frac{x-10}{11}-\frac{2(x-10)}{15}=\frac{3(x-10)}{17}-\frac{4(x-10)}{19}\)
\(\Leftrightarrow (x-10)(\frac{1}{11}+\frac{4}{19}-\frac{2}{15}-\frac{3}{17})=0\)
\(\Leftrightarrow x-10=0\Leftrightarrow x=10\)
b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)
\(\Leftrightarrow x^2-6x+9=3\)
\(\Leftrightarrow x^2-6x+6=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
a,5x2-3x+1=2x+11
\(\Leftrightarrow5x^2-3x+1-2x-11=0\)
\(\Leftrightarrow5x^2-5x-10=0\)
có a-b+c=5+5-10=0
=>\(\left\{{}\begin{matrix}x_1=-1\\x_2=2\end{matrix}\right.\)
vậy PT đã cho có 2 nghiệm là x1=-1;x2=2
b/\(\dfrac{x^2}{5}-\dfrac{2x}{3}=\dfrac{x+5}{6}\)
=>6x2-20x-5x-25=0
<=>6x2-25x-25=0
<=>(x-5)(6x+5)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\dfrac{-5}{6}\end{matrix}\right.\)
vậy PT đã cho có 2 nghiệm x1=5; x2=\(\dfrac{-5}{6}\)
c.\(\dfrac{x}{x-2}=\dfrac{10-2x}{x^2-2x}\)
=>x2+2x-10=0
\(\Delta^'=1+10=11\)
vì \(\Delta^'>0\) nên PT có 2 nghiệm phân biệt
x1=-1-\(\sqrt{11}\)
x2=-1+\(\sqrt{11}\)
d, \(\dfrac{x+0,5}{3x+1}=\dfrac{7x+2}{9x^2-1}\) ĐK x\(\ne\pm\dfrac{1}{3}\)
=>2(x+0,5)(3x-1) =2(7x+2)
=>6x2-13x-5=0
\(\Delta=169+120=289\Rightarrow\sqrt{\Delta}=17\)
vì \(\Delta\)> 0 nên PT có 2 nghiệm phân biệt
x1=\(\dfrac{13-17}{6}=\dfrac{-1}{3}\) (loại)
x2=\(\dfrac{13+17}{6}=\dfrac{5}{2}\) (thỏa mãn)
e,\(2\sqrt{3}x^2+x+1=\sqrt{3}\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{3}x^2-\left(\sqrt{3}-1\right)x+1-\sqrt{3}=0\)
\(\Delta=\left(\sqrt{3}-1\right)^2-8\sqrt{3}\left(1-\sqrt{3}\right)\)
=\(4-2\sqrt{3}-8\sqrt{3}+24\)
=25-2.5\(\sqrt{3}\)+3 =(5-\(\sqrt{3}\))2
vì \(\Delta\) >0 nên PT có 2 nghiệm phân biệt
x1=\(\dfrac{\sqrt{3}-1+5-\sqrt{3}}{4\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)
x2=\(\dfrac{\sqrt{3}-1-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\)
f/ x2+2\(\sqrt{2}\)x+4=3(x+\(\sqrt{2}\))
\(\Leftrightarrow x^2+\left(2\sqrt{2}-3\right)x+4-3\sqrt{2}=0\)
\(\Delta=8-12\sqrt{2}+9-16+12\sqrt{2}=1\)
vì \(\Delta\)>0 nên PT đã cho có 2 nghiệm phân biệt
x1=\(\dfrac{3-2\sqrt{2}+1}{2}=2-\sqrt{2}\)
x2=\(\dfrac{3-2\sqrt{2}-1}{2}=1-\sqrt{2}\)
a.
\(5x^2-3x+1=2x+11\)\(\Leftrightarrow\)\(5x^2-5x-10=0\)\(\Leftrightarrow\)\(x^2-x-2=0\)\(\Leftrightarrow\)(x-2)(x+1)=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b.
a: \(\dfrac{3\left(2x+1\right)}{20}+1>\dfrac{3x+52}{10}\)
=>\(\dfrac{6x+3}{20}+\dfrac{20}{20}>\dfrac{6x+104}{20}\)
=>6x+23>6x+104
=>23>104(sai)
vậy: \(x\in\varnothing\)
b: \(\dfrac{4x-1}{2}+\dfrac{6x-19}{6}< =\dfrac{9x-11}{3}\)
=>\(\dfrac{3\left(4x-1\right)+6x-19}{6}< =\dfrac{2\left(9x-11\right)}{6}\)
=>12x-3+6x-19<=18x-22
=>-22<=-22(luôn đúng)
Vậy: \(x\in R\)