(x-4)³=(x+4)×(x²-x-16)
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Tinh gia tri cua phan so: A= 2 x 3 + 2 x 4 x 8 + 4 x 8 x 16 + 8 x 16 x 32/ 3x 4 + 2 x 6 x 8 + 4 x 12 x 16 + 8 x 24 x 32
cmr:1-2/x-(2x+x^2/4+2x+x^2 + 2x-x^2/4-2x+x^2):(16-8x/4-2x+x^2 -16+8x/4+2x+x^2)=(x-1/x)^2
20 tháng 12 2022 lúc 20:42
Rút gọn biểu thức N=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{\sqrt{x}+16}{\sqrt{x}+2}\) với x≥0 ; x≠16
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+16}=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(\sqrt{x}+16\right)}\)
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1 x 2 x 3 x 4 x 5 x 6 x 7 x ( 4 x 8 - 16 x 2 ) x 15 x 16 x 17 ... x 2019 x 2020
Tính ngoặc tròn
\(4\cdot8-16\cdot2\)
\(=32-32\)
\(=0\)
Vậy tích trên bằng 0 ( vì có 1 thừa số = 0 )
Bài giải
1 x 2 x 3 x 4 x 5 x 6 x 7 x ( 4 x 8 - 16 x 2 ) x 15 x 16 x 17 ... x 2019 x 2020
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x ( 32 - 32 ) x 15 x 16 x 17 ... x 2019 x 2020
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 0 x 15 x 16 x 17 ... x 2019 x 2020
= 0
chắt do tui quên tính 4 x 8 - 16 x 2 nên mới vậy :v
Xem thêm câu trả lời
Rút gọn biểu thức B=(√x/√x+4 + 4/√x-4) : x+16/√x+2 ( với x lớn hơn hoặc bằng 0; x#16)
Ta có: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}+\frac{4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\right)\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{\sqrt{x}+2}{x-16}\)
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(x^2+x+1)(x^4+x^2+1)(x^8+x^4+1)(x^16+x^8+1)(x^32+x^16+1) rút gọn zùm mình với
https://www.youtube.com/watch?v=cFZDEMTQQCs
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a,1/2(x-4)-1/4(x-4/3)=2(x-1/2)
b, 3/4-(x-1/2)^2=-11/2
c,3/16+1/1/16.(x-2/3)^2=3/4
Xem chi tiết
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
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c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
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Bài 2: Tìm x, biết: a) (x+2)(x² -2x+4)-x(x²+2)=15 b) (x-2)³-(x-4)(x² + 4x+16) + 6(x+1)=49 c) (x - 1)³ + (2 - x)(4 + 2x + x²)+ 3x(x + 2) = 16 d) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 15
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
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bài 1 tìm x bt( x^2 - 4x + 16 ) ( x + 4 ) - x ( x + 1 ) ( x + 3 ) + 3x^2 0bài 2 chứng minha, ( x + 2 ) ( x - 2 ) ( x^2 + 4 ) x^4 - 16 b, ( x^2 - xy + y^2 ) ( x + y ) x^3 + y^3gúp mik với
Đọc tiếp
bài 1 tìm x bt
( x^2 - 4x + 16 ) ( x + 4 ) - x ( x + 1 ) ( x + 3 ) + 3x^2 = 0
bài 2 chứng minh
a, ( x + 2 ) ( x - 2 ) ( x^2 + 4 ) = x^4 - 16
b, ( x^2 - xy + y^2 ) ( x + y ) = x^3 + y^3
gúp mik với
Bài 2:
a: Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2-4\right)\left(x^2+4\right)\)
\(=x^4-16\)
b: Ta có:\(\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)
\(=x^3+y^3\)
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Bài 1:
Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+3\right)+3x^2=0\)
\(\Leftrightarrow x^3+64-x\left(x^2+4x+3\right)+3x^2=0\)
\(\Leftrightarrow x^3+64-x^3-4x^2-3x+3x^2=0\)
\(\Leftrightarrow-x^2-3x+64=0\)
\(\Leftrightarrow x^2+3x-64=0\)
\(\text{Δ}=3^2-4\cdot1\cdot\left(-64\right)=265\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{265}}{2}\\x_2=\dfrac{-3+\sqrt{265}}{2}\end{matrix}\right.\)
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