Giải các phương trình:
a) \(\left( {x - 7} \right)\left( {5x + 4} \right) = 0\);
b) \(\left( {2x + 9} \right)\left( {\frac{2}{3}x - 5} \right) = 0\).
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29 tháng 3 2023 lúc 20:37
Bài 6: Rút gọn các biểu thức sau:
c) C = \(\left|x-7\right|+2x-3\)
Bài 7: Giải phương trình:
a) \(\left|0,5x-5\right|=2\)
b) \(\left|5x-2\right|=-3\)
c) \(\left|\dfrac{1}{4}x+3\right|=0\)
7:
a: =>0,5x-5=2 hoặc 0,5x-5=-2
=>0,5x=3 hoặc 0,5x=7
=>x=6 hoặc x=14
b: |5x-2|=-3
mà |5x-2|>=0
nên ptvn
c: =>1/4x+3=0
=>1/4x=-3
=>x=-12
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2 tháng 2 2023 lúc 19:50
Giải các phương trình:
a) \(\left|2x\right|=x-6\)
b) \(\left|-3x\right|=x-8\)
c) \(\left|4x\right|=2x+12\)
d)\(\left|-5x\right|-16=3x\)
\(a,\left|2x\right|=x-6\)
\(\Leftrightarrow2x=x-6\)
\(\Leftrightarrow2x-x=-6\)
\(\Leftrightarrow x=-6\)
____________________
\(b,\left|-3x\right|=x-8\)
\(\Leftrightarrow3x=x-8\)
\(\Leftrightarrow3x-x=-8\)
\(\Leftrightarrow2x=-8\)
\(\Leftrightarrow x=-4\)
____________________
\(c,\left|4x\right|=2x+12\)
\(\Leftrightarrow4x=2x+12\)
\(\Leftrightarrow4x-2x=12\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
____________________
\(d,\left|-5x\right|-16=3x\)
\(\Leftrightarrow5x-16=3x\)
\(\Leftrightarrow5x-3x=16\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
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12 tháng 4 2022 lúc 21:02
Giải các phương trình:
a) \(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)
b) \(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)
c) \(1,2x^3-x^2-0,2x=0\)
a.\(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)
Ta có: \(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0;\forall x\)
\(\Rightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
b.\(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)
c.\(1,2x^3-x^2-0,2x=0\)
\(\Leftrightarrow x\left(1,2x^2-x-0,2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)
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Giải phương trình:a) dfrac{x+4}{5} - x + 4 dfrac{x}{3} - dfrac{x-2}{2}b) dfrac{4-5x}{6} dfrac{2left(-x+1right)}{2}c) dfrac{-left(x-3right)}{2} - 2 dfrac{5left(x+2right)}{4}d) dfrac{7-3x}{2} - dfrac{5+x}{5} 1
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Giải phương trình:
a) \(\dfrac{x+4}{5}\) - x + 4 = \(\dfrac{x}{3}\) - \(\dfrac{x-2}{2}\)
b) \(\dfrac{4-5x}{6}\) = \(\dfrac{2\left(-x+1\right)}{2}\)
c) \(\dfrac{-\left(x-3\right)}{2}\) - 2 = \(\dfrac{5\left(x+2\right)}{4}\)
d) \(\dfrac{7-3x}{2}\) - \(\dfrac{5+x}{5}\) = 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: \(S=\left\{\dfrac{15}{17}\right\}\)
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a) Ta có: ⇔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: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
x=−87x=−87
Vậy: 7−3x2−5+x5=17−3x2−5+x5=1
x=1517x=1517
Vậy: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
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: ⇔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 S={−87}S={−87}
d) Ta có: ⇔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 S={1517}
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GIẢI PHƯƠNG TRÌNH:
a) \(x^2-6x-4\sqrt{x^2-6x+6}=-9\)
b) \(\left(x+1\right)\left(x+4\right)=5\sqrt{x^2+5x+28}\)
b: Đặt \(x^2+5x+4=a\)
\(\Leftrightarrow a=5\sqrt{a+24}\)
\(\Leftrightarrow a^2=25a+600\)
\(\Leftrightarrow a^2-25a-600=0\)
\(\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\)
\(\Leftrightarrow a=-15\)
hay S=∅
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Bài 1. Giải các bất phương trình:a) dfrac{2x-1}{x-2} dfrac{1}{4x+2}b) left|x^2+5x+4right|x^2+3x-4c) dfrac{x+2}{3}-x+1x+3d) dfrac{3x+5}{2}-1ledfrac{x+2}{3}+xBài 2. Xét dấu các biểu thức:a) fleft(xright)left(x-3right)left(2x+3right)b) gleft(xright)left(-2x+3right)left(x-2right)left(x+4right)c) hleft(xright)dfrac{left(x+2right)left(4-xright)}{3-2x}d) kleft(xright)dfrac{2}{3-x}-dfrac{1}{3+x}
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Bài 1. Giải các bất phương trình:
a) \(\dfrac{2x-1}{x-2}< \dfrac{1}{4x+2}\)
b) \(\left|x^2+5x+4\right|>x^2+3x-4\)
c) \(\dfrac{x+2}{3}-x+1>x+3\)
d) \(\dfrac{3x+5}{2}-1\le\dfrac{x+2}{3}+x\)
Bài 2. Xét dấu các biểu thức:
a) \(f\left(x\right)=\left(x-3\right)\left(2x+3\right)\)
b) \(g\left(x\right)=\left(-2x+3\right)\left(x-2\right)\left(x+4\right)\)
c) \(h\left(x\right)=\dfrac{\left(x+2\right)\left(4-x\right)}{3-2x}\)
d) \(k\left(x\right)=\dfrac{2}{3-x}-\dfrac{1}{3+x}\)
1:
c: =>1/3x+2/3-x+1>x+3
=>-2/3x+5/3-x-3>0
=>-5/3x-4/3>0
=>-5x-4>0
=>x<-4/5
d: =>3/2x+5/2-1<=1/3x+2/3+x
=>3/2x+3/2<=4/3x+2/3
=>1/6x<=2/3-3/2=-5/6
=>x<=-5
2:
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giải các phương trình:
a)(x2+x+1)2-2x2-2x=5
b)\(\dfrac{1}{\left(x-1\right)\left(x-3\right)}\)+x2-4x+5=0
Giải các phương trình:
a) \(\left|\sin x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
b) \(\tan^2\left(x+\dfrac{\pi}{6}\right)=3\)
c) \(2\sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)
a, \(\left|sinx+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow sin^2x+sinx+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow sin^2x+sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
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b, \(tan^2\left(x+\dfrac{\pi}{6}\right)=3\)
\(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=\pm\sqrt{3}\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\pm\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
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c, \(2sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)
\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{matrix}\right.\)
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Giải các hệ phương trình:a)left{{}begin{matrix}dfrac{x}{y}dfrac{2}{3}x+y-100end{matrix}right.b)left{{}begin{matrix}left(3x+2right)left(2y-3right)6xyleft(4x+5right)left(y-5right)4xyend{matrix}right.c)left{{}begin{matrix}left(2x-3right)left(2y+4right)4xleft(y-3right)+54left(x+1right)left(3y-3right)3yleft(x+1right)-12end{matrix}right.d)left{{}begin{matrix}dfrac{2y-5x}{3}+5dfrac{y+27}{4}-2xdfrac{x+1}{3}+ydfrac{6y-5x}{7}end{matrix}right.
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Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
Giải phương trình:
a, \(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\)
b, \(\left(x-1\right)^2+3x^2=0\)
a) \(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}\right)^2-\left(x-1-\dfrac{1}{x}\right)^2=0\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}-x+1+\dfrac{1}{x}\right)\left(x+1+\dfrac{1}{x}+x-1-\dfrac{1}{x}\right)=0\\ \Leftrightarrow2\left(1+\dfrac{1}{x}\right)\cdot2x=0\\ \Leftrightarrow4x\left(1+\dfrac{1}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(S=\left\{-1;0\right\}\) là tập nghiệm của pt.
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b) Ta có: \(\left(x-1\right)^2+3x^2=0\)
\(\Leftrightarrow x^2-2x+1+3x^2=0\)
\(\Leftrightarrow4x^2-2x+1=0\)
\(\text{Δ}=\left(-2\right)^2-4\cdot4\cdot1=4-16=-12< 0\)
=> Phương trình vô nghiệm
Vậy: \(S=\varnothing\)
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