2(x-6)²-1=49
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a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
\(\left(x-1\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\ 7^{2x-6}=49\\ \Rightarrow7^{2x-6}=7^2\\ \Rightarrow2x-6=2\\ \Rightarrow2x=8\\ \Rightarrow x=4\)
3.|x+1|-2=4
3.|x+1|=4+2
3.|x+1|=6
|x+1|=6:3
|x+1|=2
Trường hợp 1 x+1=2
x=2-1
x=1
trường hợp 2
x+1=-2
x=(-2)-1
x=-3
==> x thuộc {1; -3}
k mk nha chúc học tốt
b; 36. 42 - 62.5 + 63 .62
= 36.42 - 36.5 + 63.36
= 36.(42 - 5 + 63)
= 36.100
= 3600
\(a, 3\left(2-x\right)+5\left(x-6\right)=-98\Leftrightarrow6-3x+5x-30=-98\)
\(\Leftrightarrow2x=-74\)
\(\Leftrightarrow x=-37\)
\(b,\)\(\Leftrightarrow x^2+1=0\left(v\text{ô} nghi\text{ệm}\right)ho\text{ặc} 49-x^2=0\)
\(\Leftrightarrow x=7 ho\text{ặc} x=-7\)
Chúc bạn học tốt ^_^
a)3(2-x)+5(x-6)=-98
6-3x+5x-30=98
-3x+5x =98+30-6
x(-3+5) =122
x.2 =122
x =122:2=61
b)(x^2+1)(49-x^2)=0
=>x2+1=0 hoặc 49-x2=0
x2=-1(vô lí)hoặc x2 =49
=>x=7
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
2.( x - 6)2 - 1 = 49
2.(x-6)2 = 49 + 1
2.(x-6)2 = 50
( x-6)2 = 50 : 2
(x-6)2 = 25
( x-6)2 = 52
\(\left[{}\begin{matrix}x-6=5\\x-6=-5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=11\\x=1\end{matrix}\right.\)