a) 5x-1.5=54
b)54 + (98 - x)
c)45.4x= 16
d) 3x+1.5-10:2+5 =15
e)2A + 3 = 3x Biết A=3+3+33+...+3100
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Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
1. Thực hiện phép tính bằng cách hợp lí :
a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]
= 0+(-100) = -100
b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75
= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000
c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]
= (-151) + [(-100) + 100] = -151
d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]
= 32+[(-200)+(-400)] = 32+(-600) = -568
e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65
= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)
= 65.(-70) = -4550
g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43
= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)
= 16.(-10) = -160
a) Ta có : 2 x : 2 2 = 2 5 nên x = 7.
b) Ta có: 3 x : 3 2 = 3 5 nên x = 7.
c) Ta có : 4 4 : 4 x = 4 2 nên x = 2.
d) Ta có : 5 x : 5 2 = 5 2 nên x = 4,
e) Ta có: 5 x + 1 : 5 = 5 4 nên x = 4.
f) Ta có : 4 2 x - 1 : 4 = 4 2 nên x = 2
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
A= (6x-2)^2 + (2-5x)^2+2(6x-2)(2-5x)
= (6x-2)^2 +2(6x-2)(2-5x)+ (2-5x)^2
\(=\left(6x-2+2-5x\right)^2=x^2\)
B= (2a^2+2a+1)(2a^2-2a+1)-(2a^2+1)^2
\(=\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=4a^2\)
C=(x+3)(x^2-3x+9)-(54+x^3)
\(=\left(x^3+27\right)-54-x^3=27\)
D=(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
\(=\left(2x+y\right)^3-\left(2x-y\right)^3\)
E=(a+b)^2-(a-b)^2
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a.2b=4ab\)
Secret Personv: thật.CTV lạ z
\(C=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-27-54-x^3=-81\)
a) \(2.\left|5x-3\right|-2x=14\)
\(2\left|5x-3\right|=14+2x\)
\(\left|5x-3\right|=\frac{14+2x}{2}\)
\(\Rightarrow\orbr{\begin{cases}5x-3=\frac{-14-2x}{2}\\5x-3=\frac{14+2x}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\left(5x-3\right).2=-14-2x\\\left(5x-3\right).2=14+2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}10x-6+2x=-14\\10x-6-2x=14\end{cases}\Rightarrow\orbr{\begin{cases}12x=-14+6\\8x=14+6\end{cases}}}\Rightarrow\orbr{\begin{cases}12x=-8\\8x=20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)
Những câu sau tương tự nhé.
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!