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HH
24 tháng 6 2018
Giải:
a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)
\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)
\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy ...
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)
\(\Leftrightarrow30x-37=4\)
\(\Leftrightarrow30x=41\)
\(\Leftrightarrow x=\dfrac{41}{30}\)
Vậy ...
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)
\(\Leftrightarrow x^3+8-x^3-3=14x\)
\(\Leftrightarrow5=14x\)
\(\Leftrightarrow x=\dfrac{5}{14}\)
Vậy ...
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
\(\Leftrightarrow x^3+1-x^3-3x=2\)
\(\Leftrightarrow1-3x=2\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
25 tháng 6 2018
a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)
=> \(x^2-2x-x-3x+7+9x=6\)
=> \(x^2-2x-x^2-3x+7+9x=6\)
=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)
=> \(4x=-1\)
Vậy \(x=\dfrac{-1}{4}\)
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)
=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)
=> \(42x=43\)
Vậy \(x=\dfrac{43}{42}\)
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)
=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)
=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)
=> \(5=14x\)
Vậy \(x=\dfrac{5}{14}\)
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)
=>\(\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x-3x\right)=2-1\)
=> \(-3x=1\)
Vậy \(x=\dfrac{-1}{3}\)
24 tháng 10 2019
\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)+2ab-2ac\)
\(=a\left(a-2b+2c\right)+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac\)
\(=a^2\)
24 tháng 10 2019
\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2=16\)
TC
0
22 tháng 6 2021
Bài 1 :
a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18-2\right)\)
\(=6x^2+19x-7-6x^2-x+5-16=18x-18\)
Vậy biểu thức phụ thuộc biến x
b, \(\left(x-2\right)\left(x+1\right)\left(2x+1\right)-x\left(2x^2-x-5\right)+1\)
\(=\left(x^2-x-2\right)\left(2x+1\right)-x\left(2x^2-x-5\right)+1\)
\(=2x^3+x^2-2x^2-1-4x-2-2x^3+2x+5x+1=-x^2-2+3x\)
Vậy biểu thức phụ thuộc biến x
4 tháng 9 2017
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\) =\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\) =25 c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\) =\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\) =\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\) =... =\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\) \)
d)Tương tự
4 tháng 9 2017
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
=\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\)
=\(a^2\)
b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\)
=\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\)
=25
c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\)
=\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\)
=\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\)
=...
=\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)
d)Tương tự
25 tháng 5 2022
a: \(=\left[a-\left(b-c\right)\right]^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac=a^2\)
b: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2=16\)
c: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=2^{128}-1\)
d: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{3^{64}-1}{2}\)
1 tháng 10 2016
Bài 1 :
(a^2+b^2)(x^2+y^2)=(ax+by)^2
<=> a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2
<=> a^2y^2 + b^2x^2 = 2abxy
<=> a^2y^2 + b^2x^2 - 2abxy = 0
<=> (ay - bx)^2 = 0
=> ay - bx = 0
=> ay = bx
=> a/x = b/y ( x,y khác 0)
16 tháng 10 2016
\(5x\left(x-3\right)=x-3\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)
Viết lại đề cho rõ ràng đi e, ví dụ (3x+1)mũ 2 thì viết là (3x+1)^2