7x-3.4=196
2x+3+2x=36
(52x.5x+2):25=1255
5.11x-3-3=2.11x-3
[(2x-4)+9]3=36
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1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk
\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
câu thứ 2 bạn ktra lại đề
\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
\(x^3-x^2-4=\left(x-2\right)\left(x^2+x+2\right)\)
\(x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
a) \(x^5-7x^4-x^3+43x^2-36\)
\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-6\right)\)
c) \(x^4+2x^3-15x^2-18x+64\)
\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
a: 7x+58=100
nên 7x=42
hay x=6
c: x-56:x=16
nên x-14=16
hay x=30
c)x - 56 : 4 = 16
x - 56 = 16 : 4
x- 56 = 4
x =4 + 56
x = 60
d)101 + (36 - 4x) = 105
(36- 4x ) = 105 - 101
36 - 4x = 4
4x = 36 - 4
4x = 32
x = 32:4
x = 8
a) |x-1|-2x=5
\(\Leftrightarrow\) x - 1 - 2x = 5 (đk: x\(\ge\)1 ) (1)
hoặc -x + 1 - 2x = 5 ( đk: x < 1) (2)
(1): \(\Leftrightarrow\) x - 2x = 5 + 1
\(\Leftrightarrow\) - x = 6
\(\Leftrightarrow\) x = -6 (ko thỏa mãn)
(2): \(\Leftrightarrow\) -x + 1 - 2x = 5
\(\Leftrightarrow\) -3x = 4
\(\Leftrightarrow\) x = \(\frac{-4}{3}\) (thỏa mãn)
Vậy: x = \(\frac{-4}{3}\)
b) |9-7x|=5x-3
\(\Leftrightarrow\) 9 - 7x = 5x - 3 (đk: \(\le\) \(\frac{9}{7}\) ) (1)
hoặc -9 + 7x = 5x - 3 (đk: x > \(\frac{9}{7}\) ) (2)
(1): \(\Leftrightarrow\) -7x-5x = -3-9 \(\Leftrightarrow\) -12x = -12 \(\Leftrightarrow\) x = 1 ( thỏa mãn)
(2): \(\Leftrightarrow\) 7x - 5x = -3 + 9 \(\Leftrightarrow\) 2x = 6 \(\Leftrightarrow\) x = 3 ( thỏa mãn)
Vậy: x= 1; 3
c) 5x+2 = 625
\(\Leftrightarrow\) 5x . 52 = 54
\(\Leftrightarrow\) 5x = \(\frac{5^4}{5^2}\)= 52
\(\Leftrightarrow\) x = 2
d) (2x - 3)2 = 36
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-3=6\left(đk:x\ge\frac{3}{2}\right)\text{}\Leftrightarrow x=4,5\left(TMĐK\right)\\2x-3=-6\left(đk:x< \frac{3}{2}\right)\Leftrightarrow x=-1,5\left(TMĐK\right)\end{matrix}\right.\)
Vậy: x = 4,5; -1,5
1, \(\frac{x^2+2x+1}{2x^2-2}=\frac{\left(x+1\right)^2}{2\left(x^2-1\right)}=\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{x+1}{2\left(x-1\right)}\)= \(\frac{x+1}{2x-2}\)
2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)
3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)
4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
Trả lời
Bài 1:
a)(45-25).(-11)+29.(-3-17)
=20.(-11)+29.(-20)
=20.(-11)+(-29).20
=20.[(-11)+(-29)]
=20.(-30)
=-600
b)(36-6).(-5)+21.(-17-3)
=30.(-5)+21.(-20)
=(-150)+(-420)
=-570
trả lời:
a,(45-25).(-11)+29.(-3-17)
=20.(-11)+29.(-20)
=20.(-11)+(-29).20
=20.[(-11)+(-29).20
=20.(-30)
=-600
học tốt
1, \(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
vậy \(x\) \(\in\) {-3; 3}
5, 4\(x^2\) - 36 = 0
4.(\(x^2\) - 9) = 0
\(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; 3}
7x-3.4=196
7x-3=49
x-3=2
x= -1
2x+3+2x=36
2x.9=36
2x=4
x=2
(52x.5x+2):25=1255
53x+2=517
x=5
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