Giúp mik với
a) 12x/5+x/3 < hoặc = 41/15
b) 12--2(2x+5) > 3(3-x)
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\(a,x-36:18=12-15\\ \Rightarrow x-2=-3\\ \Rightarrow x=-1\\ b,92-\left(17+x\right)=72\\ \Rightarrow17+x=20\\ \Rightarrow x=3\\ c,720:\left[41-\left(2x+5\right)\right]=40\\ \Rightarrow41-\left(2x+5\right)=18\\ \Rightarrow2x+5=23\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ d,\left(x+2\right)^3-23=41\\ \Rightarrow\left(x+2\right)^3=64\\ \Rightarrow\left(x+2\right)^3=4^3\\ \Rightarrow x+2=4\\ \Rightarrow x=2\)
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
a)\(\left(x+8\right)-11=20-15\)
\(\left(x+8\right)-11=5\)
\( x+8=5+11\)
\(x+8=16\)
\(x=8\)
b) \(2x-\left(3+x\right)=5-7\)
\(2x-\left(3+x\right)=-2\)
\(2x-3-x=-2\)
\(x=1\)
c) \( \left(3x-2^4\right)\times7^5=2\times7^6\)
\(3x-2^4=2\times\left(7^6:7^5\right) \)
\(\left(3x-2^4\right)=2\times7^2\)
\(3x-2^4=2\times49\)
\(3x-16=98\)
\(3x=114\)
\(x=38\)
Giải phương trình:
a) (x+2)3 - (x-2)3 = 12x(x-1) - 8
<=> (x2 + 3.x2.2 + 3.x.22 + 23) - (x2 - 3.x2.2 + 3.x.22 - 23) - [12x(x-1) - 8] = 0
<=> (x3 + 6x2 + 12x + 8) - (x3 - 6x2 + 12x - 8) - (12x2 - 12x - 8) = 0
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x + 8 = 0
<=> 12x +32 = 0
<=> x = \(\frac{-32}{12}\) = \(-2\frac{2}{3}\)
Vậy phương trình có nghiệm duy nhất là \(-2\frac{2}{3}\)
b) (3x-1)2 - 5(2x+1)2 + (6x-3)(2x+1) = (x-1)2
<=> (9x2 - 6x + 1) - 5(4x2 + 4x + 1) + 3(2x - 1)(2x + 1) - (x2 - 2x +1) = 0
<=> 9x2 - 6x + 1 - 20x2 - 20x - 5 + 3(4x2 - 1) - x2 + 2x -1 = 0
<=> 9x2 - 6x + 1 - 20x2 - 20x - 5 + 12x2 - 3 - x2 + 2x -1 = 0
<=> -24x - 8 = 0
<=> x = \(\frac{-8}{24}\) = \(\frac{-1}{3}\)
Vậy phương trình có nghiệm duy nhất là \(\frac{-1}{3}\)
a: =(x-y)^2+2(x-y)
=(x-y)(x-y+2)
c: =(x-3)(x+3)+(x-3)^2
=(x-3)(x+3+x-3)
=2x(x-3)
d: =(x+3)(x^2-3x+9)-4x(x+3)
=(x+3)(x^2-7x+9)
e: =(x^2-8x+7)(x^2-8x+15)-20
=(x^2-8x)^2+22(x^2-8x)+85
=(x^2-8x+17)(x^2-8x+5)
d) \(x-\left(-25+x\right)=13-x.\)
\(\Rightarrow x+25+x=13-x\)
\(\Rightarrow x+x+x=13-25\)
\(\Rightarrow3x=-12\)
\(\Rightarrow x=-4\)
e) \(15-\left(30+x\right)=x-\left(27-\text{| }-8\text{| }\right)\)
\(\Rightarrow15-30-x=x-19\)
\(\Rightarrow-15-x=x-19\)
\(\Rightarrow-15+19=x+x\)
\(\Rightarrow4=2x\)
\(x=2\)
f) \(\left(12x-4^3\right).8^3=4.8^4\)
\(\left(12x-2^6\right).2^9=2^2.2^{12}\)
\(12x-2^6=2^2.2^{12}\div2^9\)
\(12x-64=2^5=32\)
\(12x=96\)
\(x=8\)
g) \(\left[119-\left(3^3-10\right)\right].x=2448\)
\(\left[119-\left(27-10\right)\right].x=2448\)
\(\left[119-17\right].x=2448\)
\(102.x=2448\)
\(x=24\)
h)
[( 10 - x ) .2-51] : 3 - 2 = 3
[( 10 - x ) .2-51] : 3 = 3 + 2
[( 10 - x ) .2-51] : 3 = 5
( 10 - x ) .2 - 51 = 5 . 3
( 10 - x ) . 2 - 51 = 15
( 10 - x ) . 2 = 15 + 51
( 10 - x ) . 2 = 66
10 - x = 66 : 2
10 - x = 33
x = 10 - 33
x = -23
i)
(x-12)-15 = (20-7)-(18+x)
x-12-15 = 20-7-18-x
x-27 = -5-x
=> x-27+5+x = 0
2x-25 =0
2x = 0+25 = 25
k)
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
a)\(\frac{12x}{5}+\frac{x}{3}\le\frac{41}{15}\)
\(\Leftrightarrow\frac{36x}{15}+\frac{5x}{15}\le\frac{41}{15}\)
\(\Leftrightarrow\frac{36x}{15}+\frac{5x}{15}-\frac{41}{15}\le0\)
\(\Leftrightarrow\frac{36x+5x-41}{15}\le0\)
\(\Leftrightarrow31x-41\le0\)
\(\Leftrightarrow31x\le41\)
\(\Leftrightarrow x=\frac{41}{31}\)