Tìm x biết
\(\left(2x-1\right)^5-11^2=122\)
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Bài 2:
Gọi số ban đầu là \(\overline{ab}\)
Theo đề, ta có: 5a+2b=29 và 10b+a-10a-b=36
=>5a+2b=29 và -9a+9b=36
=>a=3 và b=7
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
b) 3x - 6 - (8x + 4) - (10x + 15) = 50
=> 3x - 6 - 8x - 4 - 10x - 15 = 50
=> (3x - 8x - 10x) = 6+ 4 + 15 + 50
=> -15x = 75 => x = 75 : (-15) = -5
c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)
+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3
+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1
Vậy x = 5/3 hoặc x = 1
a) (n-1)n+11-(n-1)n=0
(n-1)n(n-1)11-(n-1)n=0
(n-1)n[(n-1)11-1]=0
(n-1)n=0 hoặc (n-1)11-1=0
n-1=0 hoặc (n-1)11 =1
n=1 hoặc n-1 =1
n=1 hoặc n =2
a) 3x + 2(5 - x) = -11
=> 3x + 10 - 2x = -11
=> 3x - 2x + 10 = -11
=> x = -21
b) 3x2 - 3x(-2 + x) = 36
=> 3x2 - 3x.(-2) - 3x.x = 36
=> 3x2 + 6x - 3x2 = 36
=> 6x = 36
=> x = 6
c) x(5 - 2x) + 2x(x - 1) = 15
=> x.5 + x.(-2x) + 2x.x + 2x.(-1) = 15
=> 5x - 2x2 + 2x2 - 2x = 15
=> 3x = 15
=> x = 5
Trả lời:
a,\(3x+2.\left(5-x\right)=-11\)
\(\Leftrightarrow3x+10-2x=-11\)
\(\Leftrightarrow x+10=-11\)
\(\Leftrightarrow x=-21\)
Vậy \(x=-21\)
b,\(3x^2-3x.\left(-2+x\right)=36\)
\(\Leftrightarrow3x^2+6x-3x^2=36\)
\(\Leftrightarrow6x=36\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
c, \(x.\left(5-2x\right)+2x.\left(x-1\right)=15\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=15\)
\(\Leftrightarrow3x=15\)'
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10
\(49\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(4x^2+8x+4+4x^2-4x+1-8x^2+8-11=0\)
\(4x+2=0\)
\(4x=2\)
\(x=-\frac{1}{2}\)
<=>4(x2+2x+1)+4x2-4x+1-8x2+8-11=0
<=>4x2+8x+4+4x2-4x+1-8x2+8-11=0
<=>4x+2=0
<=>2(2x+1)=0
<=>2x+1=0
<=>x=-1/2
chuẩn
(2x-1)5-112=122 (112=121)
=> (2x-1)5 = 122 - 121
(2x-1)5 = 1
=> 2x-1 = 1
2x = 1+1
2x = 2
x = 2:2
x = 1