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\(a,\left(2x+3\right).5x=10x^2.15x\)
\(b,1011^2-1010^2=\left(1011-1010\right)\left(1011+1010\right)=2021\)
\(c,x^2+3x=x\left(x+3\right)\)
\(c,x^2+2xy-x-2y=\left(x^2-x\right)+\left(2xy-2y\right)=x\left(x-1\right)+2y\left(x-1\right)=\left(x-1\right)\left(x+2y\right)\)
a) Ta có: \(2x+x^2=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b) Ta có: \(\left(2x+1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a) \(5x+10y=5\left(x+2y\right)\)
b) \(3x^2y+9xy^2z=3xy\left(x+3yz\right)\)
g) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
h) \(x^2+9x+8=\left(x+8\right)\left(x+1\right)\)
l) \(x^2-10x+9=\left(x-1\right)\left(x-9\right)\)
k) \(x^2+x-12=\left(x+4\right)\left(x-3\right)\)
l) \(3x^2+8x+4=\left(3x+2\right)\left(x+2\right)\)
a: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x^3+2\)
\(=x^3-1-x^3+2\)
=1
b: ta có: \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)-8x^3-5\)
\(=8x^3-y^3-8x^3-5\)
\(=-y^3-5\)
c: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x^3+2\)
\(=x^3+1-x^3+2\)
=3
d: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-8x^3-5\)
\(=8x^3+y^3-8x^3-5\)
\(=y^3-5\)
a) \(\left(x-1\right)\left(x^2+x+1\right)-x^3+2=x^3-1-x^3+2=1\)
b) \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)-8x^3-5=8x^3-y^3-8x^3-5=-y^3-5\)
c) \(\left(x+1\right)\left(x^2-x+1\right)-x^3+2=x^2+1-x^3+2=3\)
d) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-8x^3-5=8x^3+y^3-8x^3-5=y^3-5\)
e) \(\left(3x+2\right)\left(9x^2-6x+4\right)-27x^3-7=27x^3+8-27x^3-7=1\)
f) \(\left(3x-2\right)\left(9x^2+6x+4\right)-27x^3-7=27x^3-8-27x^3-7=-15\)
1a. 3x2 - 2x(5 + 1,5x) + 10 = 3x2 - 10x + 3x2 = 6x2 - 10x
b. (2x - 3)(x + 7) - 2x(x+5) - x = 2x2 - 3x - 3x - 21 - 2x2 - 10x = -16x - 21
c. (x + 3)(x - 3) - (x - 5)(x + 2) = x2 - 32 - (x2 + 2x - 5x - 10) = x2 - 9 - x2 - 2x + 5x + 10 = 3x + 1
d. (x - 1)2 - (x + 2)(x - 2) = x2 - 2x + 1 - (x2 - 22) = x2 - 2x + 1 - x2 + 4 = 2x + 5
e. (x + 2)(x2 - 2x + 4) - x(x2 + 2) = x3 + 23 - x3 - 2x = 8 - 2x
f. (x + 2y)3 - 6xy(x + 2y) = x3 + 6x2y + 12xy2 + 8y - 6x2y + 12xy2 = x2 + 24xy2 + 8y
Bài 3:
a: \(x^3+x^2-4x-4=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
b: \(x^2-4x+4-y^2=\left(x-2-y\right)\left(x-2+y\right)\)
c: \(\left(x^2+9\right)^2-36x^2=\left(x+3\right)^2\cdot\left(x-3\right)^2\)
Em cần giúp câu nào hả em? Em nên chụp 1-2 ý cho 1 lần hỏi nhá, như thế mọi người sẽ dễ dàng giúp em hơn
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a, \(3x-4=-x+8\)
\(< =>3x+x=8+4\)
\(< =>4x=12\)
\(< =>x=\frac{12}{4}=3\)
b, \(\frac{2x+1}{6}+\frac{x-7}{12}=10\)
\(< =>\frac{2\left(2x+1\right)}{12}+\frac{x-7}{12}=\frac{120}{12}\)
\(< =>4x+2+x-7=120\)
\(< =>5x=120+5=125\)
\(< =>x=\frac{125}{5}=\frac{5^3}{5}=5^2=25\)
a, Xét ΔBHA và ΔBAC có :
\(\widehat{A}=\widehat{H}=90^0\)
\(\widehat{B}:chung\)
\(\Rightarrow\Delta BHA\sim\Delta BAC\left(g-g\right)\)
b, Xét ΔCHA và ΔCAB có :
\(\widehat{A}=\widehat{H}=90^0\)
\(\widehat{C}:chung\)
\(\Rightarrow\Delta CHA\sim\Delta CAB\left(g-g\right)\)
c, Xét ΔAHB và ΔCHA có :
\(\widehat{BHA}=\widehat{CHA}=90^0\)
\(\widehat{B}=\widehat{HAC}\left(phụ\cdot với\cdot\widehat{C}\right)\)
\(\Rightarrow\Delta AHB\sim\Delta CHA\left(g-g\right)\)