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a) x = 0
b) x = -1
c) x = -9
d) x = 24
e) x = 2 hoặc x = -4
f) x = 5 hoặc x = -3
x + 1 3 + 3 2 x + 1 4 = 2 x + 3 x + 1 6 + 7 + 12 x 12 ⇔ x + 1 3 + 6 x + 3 4 = 5 x + 3 6 + 7 + 12 x 12
⇔ 4(x + 1) + 3(6x + 3) = 2(5x + 3) + 7 + 12x
⇔ 4x + 4 + 18x + 9 = 10x + 6 + 7 + 12x
⇔ 4x + 18x – 10x – 12x = 6 + 7 – 4 – 9
⇔ 0x = 0
Phương trình có vô số nghiệm.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a) \(\sqrt[]{x^2-2x+4}=2x-2\)
\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)
Giải pt \(3x^2-6x=0\)
\(\Leftrightarrow3x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=2\)
c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
A,-16+23+x=-16
<=>x=-16+16-23
<=>x=-23
B,10-2(4-3x)=-4
<=>10-8+6x=-4
<=>6x=-4-10+8
<=>6x=-6
<=>x=-1
C,-12+3(-x+7)=-18
<=>-12-3x+21=-18
<=>-3x=-18+12-21
<=>-3x=-27
<=>x=9
D,24:(3x-2)=-3
<=>24/(3x-2)=-3
<=>-3(3x-2)=24
<=>3x-2=-8
<=>3x=-6
<=>x=-2
E,|x+8|-7=8
<=>|x+8|=15
<=>x+8=+-15
Chia 2 TH:
TH1:x+8=15
<=>x=7
TH2:x+8=-15
<=>x=-23
F,-45:5(-3-2x)=3
<=>-9(-3-2x)=3
<=>27+18x=3
<=>18x=-24
<=>x=-4/3
G,|x-1|=0
<=>x-1=0
<=>x=1
H,-13|x|=-26
<=>|x|=2
<=>x=+-2
a) -16 + 23 + x = -16
x = -16 + 16 - 23
x = -23
Vậy x = -23
b) 10 - 2. (4 - 3x) = -4
10 - 2 . 4 + 2. 3x = -4
10 - 8 + 6x = -4
6x = -4 - 10 + 8
6x = -6
x = (-6) : 6
x = -1
Vậy x = -1
c) -12 + 3. (-x + 7) = -18
-12 + 3. (-x) + 3. 7 = -18
-12 + (-3x) + 21 = -18
-3x = -18 + 12 - 21
-3x = -27
x = (-27) : (-3)
x = -9
Vậy x = -9
d) 24 : (3x - 2) = -3
3x - 2 = 24 : (-3)
3x - 2 = -8
3x = -8 + 2
3x = -6
x = (-6) : 3
x = -2
Vậy x = -2
e) lx + 8l - 7 = 8
lx + 8l = 8 + 7
lx + 8l = 15
\(\Rightarrow\hept{\begin{cases}x+8=15\\x=18-5\\x=13\end{cases}hay\hept{\begin{cases}x+8=-15\\x=-15-8\\x=-23\end{cases}}}\)
Vậy \(x\in\left\{13;-23\right\}\)
f) (-45) : 5. (-3 - 2x) = 3
(-9). (-3 - 2x) = 3
(-9). (-3) + 9. 2x = 3
27 + 18x = 3
18x = 3 - 27
18x = -24
x = (-24) : 18
x =\(\frac{-4}{3}\)
Vậy x =\(\frac{-4}{3}\)
g) lx - 1l = 0
\(\Rightarrow x-1=0\)
x = 0 + 1
x = 1
Vậy x = 1
h) (-13). lxl = -26
lxl = (-26) : (-13)
lxl = 2
\(\Rightarrow x=2\)
Vậy x = 2
Chúc bạn học tốt!!!
b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-7;-9;-3;-13\right\}\)
thanks
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