(2x + 9).3=54
8 -(7-x)=3
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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
a, \(\left(2x-1\right)\left(x+3\right)-2x^2+5x=7\)
\(\Leftrightarrow2x^2+6x-x-3-2x^2+5x=7\)
\(\Leftrightarrow2x^2+5x-3-2x^2+5x=7\)
\(\Leftrightarrow10x-10=0\Leftrightarrow x=1\)
b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-4\right)\left(x+4\right)=54\)
\(\Leftrightarrow\left(x^3+27\right)-x\left(x^2-16\right)=54\)
\(\Leftrightarrow x^3+27-x^3+16x=54\)
\(\Leftrightarrow-27+16x=0\Leftrightarrow x=\frac{27}{16}\)
Bài 4:
a: =>7/x-5=2
=>x-5=7/2
=>x=17/2
b: =>1-2x=-5
=>2x=6
=>x=3
c: =>2x-3=5 hoặc 2x-3=-5
=>2x=8 hoặc 2x=-2
=>x=-1 hoặc x=4
d: =>2(x+1)^2+17=21
=>2(x+1)^2=4
=>(x+1)^2=2
=>\(x+1=\pm\sqrt{2}\)
=>\(x=\pm\sqrt{2}-1\)
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
\(\text{(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)+(x+9)=54}\)
\(\text{x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9=54}\)
\(9x+\left(1+2+3+4+5+6+7+8+9\right)=54\)
\(9x-45=54\)
\(9x=99\)
\(x=11\)
\(\text{(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)+(x+9)=54}\)
\(\text{x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9=54}\)
\(\text{9x + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 54 }\)
\(\text{9x − 45 = 54 }\)
\(\text{9x = 99 }\)
\(\text{x = 11}\)
9 x 6 = 54 9 x 7 = 63 9 x 5 = 45 9 x 8 = 72
54 : 9 = 6 63 : 9 = 7 45 : 9 = 5 72 : 9 = 8
54 : 6 = 9 63 : 7 = 9 45 : 5 = 9 72 : 8 = 9
9 x 6 = 54 9 x 7 = 63 9 x 5 = 45 9 x 8 = 72
54 : 9 = 6 63 : 9 = 7 45 : 9 = 5 72 : 9 = 8
54 : 6 = 9 63 : 7 = 9 45 : 5 =9 72 : 8 =9
a)(2x +9).3=54 b)8-(7-x) =3
2.x +9 =54:3 7-x =8-3
2.x +9 =18 7-x =5
2.x =18-9 x =7-5
2.x =9 x = 2
x =9:2 vậy x=2
x = 4,5
vậy x =4,5