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a) \(4x-3=11-3x\)
\(\Leftrightarrow4x+3x=11+3\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
Vậy .............
b) \(x^3-4x^2+3x=0\)
\(\Leftrightarrow x\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x\left(x^2-x-3x+3\right)=0\)
\(\Leftrightarrow x\left[x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
Vậy .................
P/s: câu c bn gõ lại dc ko
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15-8x=9-5x
<=> 5x - 8x = 9 - 15
<=> -3x = -6
<=> x = 2
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a. 7x+12= 0 \(\Leftrightarrow7x=-12\Leftrightarrow x=-\frac{12}{7}\)
b.-2x+14=0 \(\Leftrightarrow-2x=-14\Leftrightarrow x=7\)
c. 3x+1=7x-11 \(\Leftrightarrow3x-7x=-11-1\Leftrightarrow-4x=-12\Leftrightarrow x=3\)
d.2x+x+12=0 \(\Leftrightarrow2x+x=-12\Leftrightarrow3x=-12\Leftrightarrow x=-4\)
e.x-5=3-x \(\Leftrightarrow x+x=3+5\Leftrightarrow2x=8\Leftrightarrow x=4\)
f. 7-3x=9-x \(\Leftrightarrow-3x+x=9-7\Leftrightarrow-2x=2\Leftrightarrow x=-1\)
g. 8-3x=6x+7 \(\Leftrightarrow-3x-6x=7-8\Leftrightarrow-9x=-1\Leftrightarrow x=\frac{1}{9}\)
h. 11-2x=x-1\(\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=4\)
k. 15-8x=9-5x \(\Leftrightarrow-8x+5x=9-15\Leftrightarrow-3x=-6\Leftrightarrow x=2\)
l. 3+2x=5+2 \(\Leftrightarrow2x=5+2-3\Leftrightarrow2x=4\Leftrightarrow x=2\)
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20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
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Bài 1:
a) Ta có: 7x+12=0
\(\Leftrightarrow7x=-12\)
hay \(x=-\frac{12}{7}\)
Vậy: \(x=-\frac{12}{7}\)
b) Ta có: 5x-2=0
\(\Leftrightarrow5x=2\)
hay \(x=\frac{2}{5}\)
Vậy: \(x=\frac{2}{5}\)
c) Ta có: 12-6x=0
\(\Leftrightarrow6x=12\)
hay x=2
Vậy: x=2
d) Ta có: -2x+14=0
⇔-2x=-14
hay x=7
Vậy: x=7
Bài 2:
a) Ta có: 3x+1=7x-11
⇔3x+1-7x+11=0
⇔-4x+12=0
⇔-4x=-12
hay x=3
Vậy: x=3
b) Ta có: 2x+x+12=0
⇔3x+12=0
⇔3x=-12
hay x=-4
Vậy: x=-4
c) Ta có: x-5=3-x
⇔x-5-3+x=0
⇔2x-8=0
⇔2x=8
hay x=4
Vậy: x=4
d) Ta có: 7-3x=9-x
⇔7-3x-9+x=0
⇔-2x-2=0
⇔-2x=2
hay x=-1
Vậy: x=-1
e) Ta có: 5-3x=6x+7
⇔5-3x-6x-7=0
⇔-9x-2=0
⇔-9x=2
hay \(x=\frac{-2}{9}\)
Vậy: \(x=\frac{-2}{9}\)
f) Ta có: 11-2x=x-1
⇔11-2x-x+1=0
⇔12-3x=0
⇔3x=12
hay x=4
Vậy: x=4
g) Ta có: 15-8x=9-5
⇔15-8x=4
⇔8x=11
hay \(x=\frac{11}{8}\)
Vậy: \(x=\frac{11}{8}\)
Bài 3:
a) Ta có: 0,25x+1,5=0
⇔0,25x=-1,5
hay x=-6
Vậy: x=-6
b) Ta có: 6,36-5,2x=0
⇔5,2x=6,36
hay \(x=\frac{159}{130}\)
Vậy: \(x=\frac{159}{130}\)
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\(\left|5-7x\right|=\dfrac{1}{4}\)\(\Rightarrow\left\{{}\begin{matrix}5-7x=\dfrac{1}{4}\\5-7x=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}7x=\dfrac{19}{4}\\7x=\dfrac{21}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{28}\\x=\dfrac{3}{4}\end{matrix}\right.\)\(\left|4x-11\right|=\dfrac{1}{2}x-1\left\{{}\begin{matrix}4x-11=\dfrac{1}{2}x-1\\4x-11=-\left(\dfrac{1}{2}x-1\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x-\dfrac{1}{2}x=11-1\\4x-11=-\dfrac{1}{2}x+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\left(4-\dfrac{1}{2}\right)=10\\4x+\dfrac{1}{2}x=11+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\times\dfrac{7}{2}=10\\x\left(4+\dfrac{1}{2}\right)=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{7}\\x\times\dfrac{9}{2}=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{7}\\x=\dfrac{8}{3}\end{matrix}\right.\)
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\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}< \frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}< \frac{7x^2-14x-5}{15}\)
\(\Rightarrow3\left(2x+1\right)^2-5\left(x-1\right)^2< 7x^2-14x-5\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)< 7x^2-14x-5\)
\(\Leftrightarrow\left(12x^2+12x+3\right)-\left(5x^2-10x+5\right)< 7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5< 7x^2-14x-5\)
\(\Leftrightarrow7x^2+22x-2< 7x^2-14x-5\)
\(\Leftrightarrow7x^2+22x-2-7x^2+14x+5< 0\)
\(\Leftrightarrow36x+3< 0\)
\(\Leftrightarrow36x< -3\)
\(\Leftrightarrow x< -\frac{3}{36}\)
\(\Leftrightarrow x< -\frac{1}{12}\)
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a. -3x^2+7x+10=0
⇔ 3x^2-7x-10=0
⇔ 3x^2+3x-10x-10=0
⇔ 3x(x+1)-10(x+1)=0
⇔ (3x-10)(x+1)=0
⇔ 3x-10=0 hoặc x+1=0
⇔ x=10/3 hặc x=-1
1-2x<=7x-11/5
<=>-2x-7x<=-1-11/5
<=>-9x<=-16/5
<=> x >=16/45
chúc bạn học tốt ^_^