Question

[Oz Vessalius]

1) Tìm x biết: -4x(x-5)-2x(8-2x) = -3

2) Tìm x, y, z biết:

x³/8 = y³/64 = z³/216 và x² + y² + z² = 14

3) Cho x₁ + x₂ + x₃ + x₄ +.......+ x₅₀ + x₅₁ = 0

và x₁ + x₂ = x₃ + x₄ = x₅ + x₆ = ...... = x₄₉ + x₅₀ = 1. Tính x₅₀.

CÁc bn giúp mk vs!!!!! mai phải nộp rồi /*HELP ME*/

Answer 1

[Mặc Chinh Vũ]

\(1)-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)

\(\Rightarrow-4x^2-\left(-20x\right)-16x+4x^2=-3\)

\(\Rightarrow20x-14x=-3\)

\(\Rightarrow6x=-3\)

\(\Rightarrow x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\)

\(2)\) Theo bài ra, ta có: \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(x^2+y^2+z^2=14\)

\(\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)

\(\Rightarrow\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)

\(\Rightarrow\sqrt[3]{\left(\dfrac{x}{2}\right)^3}=\sqrt[3]{\left(\dfrac{y}{4}\right)^3}=\sqrt[3]{\left(\dfrac{z}{6}\right)^3}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{6}\right)^2\)

\(\Rightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)

Suy ra:

\(+)\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}.4=1=\left(\pm1\right)^2\Rightarrow x=\pm1\)

\(+)\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{16}.4=\dfrac{1}{4}=\left(\pm\dfrac{1}{2}\right)^2\Rightarrow y=\pm\dfrac{1}{2}\)

\(+)\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{36}.4=\dfrac{1}{9}=\left(\pm\dfrac{1}{3}\right)^2\Rightarrow z=\pm\dfrac{1}{3}\)

Vậy \(\left(x;y;z\right)\in\left\{\left(-1;-\dfrac{1}{2};-\dfrac{1}{3}\right);\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\right\}\)

Answer 2

[Mặc Chinh Vũ]

Oz Vessalius Câu 3 bạn xem lại xem có sai đề không?