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sai. Vì động năng là năng lượng của khối lượng đg chuyển động
bn hok tốt nhé!!!!!!!!!
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
a ) \(\left(2x-1\right)^2+\left(x+4\right)^2+5=2x\left(x+1\right)+\left(x+2\right)^2+2x^2-2x+18\)
\(\Leftrightarrow4x^2-4x+1+x^2+8x+16+5=2x^2+2x+x^2+4x+4+2x^2-2x+18\)
\(\Leftrightarrow5x^2+4x+22=5x^2+4x+22\)
=> PT có vô số nghiệm
b ) \(\dfrac{5x-7}{4}-\dfrac{9x-4}{5}=-x-\dfrac{19-9x}{20}\)
\(\Leftrightarrow\dfrac{25x-35-36x+16}{20}=\dfrac{-20x-19+9x}{20}\)
\(\Leftrightarrow\dfrac{-11x-19}{20}=\dfrac{-11x-19}{20}\)
=> PT có vô số nghiệm
c ) \(\left|y-3\right|=y-3\)
TH 1 : \(y\ge3\)
\(\Rightarrow y-3\ge0\Rightarrow\left|y-3\right|=y-3\)
Do \(\left|y-3\right|=y-3\)
\(\Rightarrow y-3=y-3\)
Nên : \(y\ge3\) , PT vô số nghiệm
TH 2 : \(y< 3\Rightarrow y-3< 0\Rightarrow\left|y-3\right|=3-y\)
Do \(\left|y-3\right|=y-3\)
\(\Rightarrow3-y=y-3\)
\(\Rightarrow3-y-y+3=0\)
\(\Rightarrow6-2y=0\)
\(\Rightarrow y=3\) ( L ; do y < 3 )
Vậy \(y\ge3\) thì PT vô số nghiệm
Hình như bạn ghi thiếu dấu + đó
Bạn áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\)
Khi đó\(a=x^2+1\)
\(b=x^2+6x-1\)
a) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+5x=30-144\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: S={6}
b) Ta có: \(\dfrac{4-5x}{6}=\dfrac{2\left(-x+1\right)}{2}\)
\(\Leftrightarrow2\cdot\left(4-5x\right)=12\left(-x+1\right)\)
\(\Leftrightarrow2-10x=-12x+12\)
\(\Leftrightarrow2-10x+12x-12=0\)
\(\Leftrightarrow2x-10=0\)
\(\Leftrightarrow2x=10\)
hay x=5
Vậy: S={5}
c) Ta có: \(\dfrac{-\left(x-3\right)}{2}-2=\dfrac{5\left(x+2\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3-x\right)}{4}-\dfrac{8}{4}=\dfrac{5\left(x+2\right)}{4}\)
\(\Leftrightarrow6-2x-8=5x+10\)
\(\Leftrightarrow-2x+2-5x-10=0\)
\(\Leftrightarrow-7x-8=0\)
\(\Leftrightarrow-7x=8\)
hay \(x=-\dfrac{8}{7}\)
Vậy: \(S=\left\{-\dfrac{8}{7}\right\}\)
d) Ta có: \(\dfrac{7-3x}{2}-\dfrac{5+x}{5}=1\)
\(\Leftrightarrow\dfrac{5\left(7-3x\right)}{10}-\dfrac{2\left(x+5\right)}{10}=\dfrac{10}{10}\)
\(\Leftrightarrow35-15x-2x-10-10=0\)
\(\Leftrightarrow-17x+15=0\)
\(\Leftrightarrow-17x=-15\)
hay \(x=\dfrac{15}{17}\)
Vậy: \(S=\left\{\dfrac{15}{17}\right\}\)
a) Ta có: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+5x=30−144⇔−24x+5x=30−144
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: S={6}
b) Ta có: 4−5x6=2(−x+1)24−5x6=2(−x+1)2
⇔2⋅(4−5x)=12(−x+1)⇔2⋅(4−5x)=12(−x+1)
⇔2−10x=−12x+12⇔2−10x=−12x+12
⇔2−10x+12x−12=0⇔2−10x+12x−12=0
⇔2x−10=0⇔2x−10=0
⇔2x=10⇔2x=10
hay x=5
Vậy: S={5}
c) Ta có: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4
⇔6−2x−8=5x+10⇔6−2x−8=5x+10
⇔−2x+2−5x−10=0⇔−2x+2−5x−10=0
⇔−7x−8=0⇔−7x−8=0
⇔−7x=8⇔−7x=8
hay x=−87x=−87
Vậy: S={−87}S={−87}
d) Ta có: 7−3x2−5+x5=17−3x2−5+x5=1
⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010
⇔35−15x−2x−10−10=0⇔35−15x−2x−10−10=0
⇔−17x+15=0⇔−17x+15=0
⇔−17x=−15⇔−17x=−15
hay x=1517x=1517
Vậy: S={1517}
a) Ta có: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+5x=30−144⇔−24x+5x=30−144
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: S={6}
b) Ta có: 4−5x6=2(−x+1)24−5x6=2(−x+1)2
⇔2⋅(4−5x)=12(−x+1)⇔2⋅(4−5x)=12(−x+1)
⇔2−10x=−12x+12⇔2−10x=−12x+12
⇔2−10x+12x−12=0⇔2−10x+12x−12=0
⇔2x−10=0⇔2x−10=0
⇔2x=10⇔2x=10
hay x=5
Vậy: S={5}
c) Ta có: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4
⇔6−2x−8=5x+10⇔6−2x−8=5x+10
⇔−2x+2−5x−10=0⇔−2x+2−5x−10=0
⇔−7x−8=0⇔−7x−8=0
⇔−7x=8⇔−7x=8
hay x=−87x=−87
Vậy: S={−87}S={−87}
d) Ta có: 7−3x2−5+x5=17−3x2−5+x5=1
⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010
⇔35−15x−2x−10−10=0⇔35−15x−2x−10−10=0
⇔−17x+15=0⇔−17x+15=0
⇔−17x=−15⇔−17x=−15
hay x=1517x=1517
Vậy: S={1517}
\(\dfrac{-5x+7-2x+3-3x+4}{2-x}=\dfrac{-10x+14}{2-x}=\dfrac{-2\left(5x-7\right)}{2-x}=\dfrac{2\left(5x-7\right)}{x-2}.\)