120 : [56 - (3x - 1) = 2^3. 5
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tìm x
a) (3x + 5) + (2x + 10) =100
=> (3x + 2x) + ( 5 + 10 ) = 100
=> 5x + 15 =100
=> 5x = 100-15
=> 5x = 85
=> x =85:5
=> x = 17
b) 10x - 24-8x=56
=> (10x - 8x) -24 = 56
=> 2x - 24 = 56
=> 2x = 56 - 24
=> 2x = 32
=> x = 32 : 2
=> x = 16
c) 2(x+1)+(x+4) =120
=> 2.5x = 120
=> 5x = 120 : 2
=> 5x = 60
=> x = 60:5
=> x = 12
có gì sai sót mong em thông cảm !!!
Anh giải theo lập phương trình, mong e thông cảm. Giờ a giải cách lớp 6 cho nhé:
1.
10x -24- 8x= 56
(10x-8x) -24= 56
2x -24= 56
2x= 56- 24
2x= 32
x= 32/2
x=16
[(6x - 72) : 2 - 84]. 28 = 5628
=> (6x - 72) : 2 - 84 = 5628 : 28 = 201
=> (6x - 72) : 2 = 201 + 84 = 285
=> 6x - 72 = 285 . 2 = 570
=> 6x = 570 + 72 = 642
=> x = 642 : 6
=> x = 107
a) \(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)
\(\Rightarrow15x-9x^2-25+15x+9\left(x^2+2x+1\right)-30=0\)
\(\Rightarrow30x-9x^2-25+9x^2+18x+9-30=0\)
\(\Rightarrow48x-46=0\)
\(\Rightarrow x=\frac{23}{24}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Rightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow8x+17=16\)
\(\Rightarrow8x=-1\)
\(\Rightarrow x=\frac{-1}{8}\)
c) \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)
\(\Rightarrow\left(y-2\right)^3-\left(y^3-3^3\right)+6\left(y^2+2y+1\right)=49\)
\(\Rightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)
\(\Rightarrow\left(y^3-y^3\right)+\left(-6y^2+6y^2\right)+\left(12y+12y\right)+\left(-8+27+6\right)=49\)
\(\Rightarrow24y+25=49\)
\(\Rightarrow24y=24\)
\(\Rightarrow y=1\)
d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)
\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)
\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)
\(\Rightarrow3y^2+12y+13=28\)
\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)
\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)
\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
a) Ta có 2x + 2x + 1 + 2x + 2 = 56
⇒ 2x ( 1 + 21 + 22 ) = 56
⇒ 2x . 7 = 56
⇒ 2x = 56 : 7 = 8 = 23
Vậy x = 3
b) Ta có 3x + 3x + 2 + 3x + 3 = 111
⇒ 3x ( 1 + 32 + 33 ) = 111
⇒ 3x . 37 = 111
⇒ 3x = 111 : 37 = 3 = 31
Vậy x = 1
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
a,91-5(5+x)=61
=>25+5x=91-61=30
=>5x=30-25=5
=>x=1
b,\([\left(x+34\right)-50]\)x2=56
=>(x+34)-50=56:2=28
=>x+34=28+50=78
=>x=78-34=44.
c,1045-\([2015-\left(3x-24\right)]\)=5
=>2015-(3x-24)=1045-5=1040
=>3x-24=2015-1040=975
=>3x=975+24=999
=>x=999:3=333
d,\([195-\left(3x-27\right)]\)x39=4212
=>195-(3x-27)=4212:39=108
=>3x-27=195-108=87
=>3x=87+27=117
=>x=39
e,30-3(x-2)=18
=>30-3x+6=18
=>30-3x=18-6=12
=>3x=30-12=18
=>x=18:3=6
a) \(...\Rightarrow5\left(5+x\right)=91-61=30\)
\(\Rightarrow\left(5+x\right)=30:5=6\Rightarrow x=6-5=1\)
b) \(...\Rightarrow\left(x+34\right)-50=56:2=28\)
\(\Rightarrow\left(x+34\right)=28+50=78\Rightarrow x=78-34=44\)
c) \(...\Rightarrow2015-\left(3x-24\right)=1045-5=1040\)
\(\Rightarrow\left(3x-24\right)=2015-1040=975\)
\(\Rightarrow3x=975+24=999\Rightarrow x=999:3=333\)
d) \(...\Rightarrow195-\left(3x-27\right)=4212:39=108\)
\(\Rightarrow\left(3x-27\right)=195-108=87\)
\(\Rightarrow3x=87+27=114\Rightarrow x=114:3=38\)
e) \(...\Rightarrow3\left(x-2\right)=30-18=12\Rightarrow x-2=12:3=4\)
\(\Rightarrow x=4+2=6\)
=120:[56 - (3x - 1)]=40
=> 56-(3x-1)= 120:40
=> 56-(3x-1)=3
=> (3x-1)=56-3
=> (3x-1)= 53
=> 3x-1 = 53
=> 3x = 53 +1
=> x= 54 :3
=> x = 18 Vậy tập nghiệm là {18}
18 nha