tìm x biết:
a,3x+3 _ 3x+2 _ 3x =153
b,(1/2x -3)2 =9
c,1/32.3x.34=37
d,(7x+2)-1=3-2
ai làm nhanh bài này mk tick cho nà ^^
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
Bài 1:
Ta có:
\(y-x=25\Rightarrow y=25+x\)
Mà \(7x=4y\Rightarrow7x=4\cdot\left(25+x\right)\)
\(7x=100+4x\)
\(\Rightarrow7x-4x=100\)
\(3x=100\)
\(x=\frac{100}{3}\)
bài 1 :
Ta có: 7x=4y ⇔ x/4=y/7
áp dụng tính chất dãy tỉ số bằng nhau ta có
x/4=y/7=(y-x)/(7-4)=100/3
⇒x= 4 x 100/3=400/3 ; y = 7 x 100/3=700/3
bài 2
ta có x/5 = y/6 ⇔ x/20=y/24
y/8 = z/7 ⇔ y/24=z/21
⇒x/20=y/24=z/21
ADTCDTSBN(bài 1 có)
x/20=y/24=z/21=(x+y)/(20+24)=69/48=23/16
⇒x= 20 x 23/16 = 115/4
y= 24x 23/16=138/2
z=21x23/16=483/16
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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
bài a mình ko biết làm
bài b;(\(\frac{1}{2}.x-3=\)3
1/2x=6
x=3
bài c:\(\frac{1}{3}^2.3^x\cdot3^4=3^7\)
=>\(\frac{1}{3}^2\cdot3^4=\frac{3^7}{3^x}\)
=>\(\frac{3^7}{3^x}=9\)
vậy x=3^7/3^2=3^5
vậy x = 5
d)\(\left(7x+2\right)^{-1}=9^{-1}\)
=>7x+2=9
vậy 7x=9-2
7x=7
x=1
bài b hình như bn làm sai2 ý tất cả mũ 2 sao mk ko thấy mũ 2 ở đâu