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Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)
Do đó: x=54; y=36
Bài 4:
a, F(\(x\)) = m\(x\) + 3 có nghiệm \(x\) = 2
⇔ F(2) = 0 ⇔ m.2 + 3 = 0
2m = -3
m = - \(\dfrac{3}{2}\)
b, F(\(x\)) = m\(x\) - 5 có nghiệm \(x\) = 3 ⇔ F(3) = 0
⇔3m - 5 = 0 ⇒ m = \(\dfrac{5}{3}\)
c, F(\(x\)) = \(x^2\) + a\(x\) + b có 2 nghiệm phân biệt \(x\) = 1; \(x\) = 0
⇔ \(\left\{{}\begin{matrix}0+0+b=0\\1+a+b=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}b=0\\a=-1\end{matrix}\right.\)
1.
a//b mà b⊥CD nên a⊥CD
Do đó \(\widehat{D}=90^0\)
Góc A là góc nào??
2.
a, Vì a và b cùng vuông góc với MN nên a//b
b, a//b \(\Rightarrow\widehat{P}+\widehat{Q}=180^0\left(trong.cùng.phía\right)\Rightarrow\widehat{P}=70^0\)
Bài 6
a) (3x² + 5) + [(2x² - 5x) - (5x² + 4)]
= 3x² + 5 + (2x² - 5x - 5x² - 4)
= 3x² + 5 + 2x² - 5x - 5x² - 4
= (3x² + 2x² - 5x²) - 5x + (5 - 4)
= -5x + 1
---------‐----------
b) (x + 2)(x² - 2x + 4)
= x.x² - x.2x + x.4 + 2.x² - 2.2x + 2.4
= x³ - 2x² + 4x + 2x² - 4x + 8
= x³ + (-2x² + 2x²) + (4x - 4x) + 8
= x³ + 8
-------------------
c) (4x³ - 8x² + 13x - 5) : (2x - 1)
= (4x³ - 2x² - 6x² + 3x + 10x - 5) : (2x - 1)
= [(4x³ - 2x²) - (6x² - 3x) + (10x - 5)] : (2x - 1)
= [2x²(2x - 1) - 3x(2x - 1) + 5(2x - 1)] : (2x - 1)
= (2x - 1)(2x² - 3x + 5) : (2x - 1)
= 2x² - 3x + 5
Bài 4:
a: \(\sqrt{6^2+8^2}-3\sqrt{25}\)
\(=\sqrt{36+64}-3\cdot5\)
\(=\sqrt{100}-15=10-15=-5\)
b: \(\left(-5\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)-\sqrt{\dfrac{4}{3^2}}\cdot\left(-\dfrac{2}{3}\right)\)
\(=\left(-\dfrac{11}{2}\right)\left(-\dfrac{1}{2}\right)-\dfrac{2}{3}\cdot\left(-\dfrac{2}{3}\right)\)
\(=\dfrac{11}{4}+\dfrac{4}{9}=\dfrac{99+16}{36}=\dfrac{115}{36}\)
c: \(\sqrt{16}\cdot\sqrt{4}-\sqrt{25}+2\sqrt{49}\)
\(=4\cdot2-5+2\cdot7\)
\(=8-5+14=22-5=17\)
d: \(\dfrac{1}{\sqrt{36}}+\dfrac{\sqrt{25}}{6}-\sqrt{0,81}\)
\(=\dfrac{1}{6}+\dfrac{5}{6}-0,9\)
=1-0,9
=0,1
e: \(-\dfrac{\sqrt{9}}{16}+\dfrac{5}{\sqrt{36}}\)
\(=-\dfrac{3}{16}+\dfrac{5}{6}\)
\(=\dfrac{-9+40}{48}=\dfrac{31}{48}\)
f: \(\dfrac{\sqrt{9}}{8}\cdot\dfrac{16}{\sqrt{225}}-\dfrac{3}{4\sqrt{4}}\cdot\dfrac{2}{5\sqrt{3^2}}\)
\(=\dfrac{3}{8}\cdot\dfrac{16}{15}-\dfrac{3}{4\cdot2}\cdot\dfrac{2}{5\cdot3}\)
\(=\dfrac{3}{15}\cdot\dfrac{16}{8}-\dfrac{1}{20}\)
\(=\dfrac{1}{5}\cdot2-\dfrac{1}{20}=\dfrac{2}{5}-\dfrac{1}{20}=\dfrac{7}{20}\)
Bài 4:
a) \(\sqrt{6^2+8^2}-3\sqrt{25}\)
\(=\sqrt{36+64}-3\cdot5\)
\(=\sqrt{100}-15\)
\(=10-15\)
\(=-5\)
b) \(\left(-5\dfrac{1}{2}\right)\cdot\left(-\dfrac{1}{2}\right)-\sqrt{\dfrac{4}{3^2}}\cdot\left(-\dfrac{2}{3}\right)\)
\(=\left(-\dfrac{11}{2}\right)\cdot\left(-\dfrac{1}{2}\right)-\dfrac{\sqrt{4}}{\sqrt{3^2}}\cdot\left(-\dfrac{2}{3}\right)\)
\(=\dfrac{11}{4}-\dfrac{2}{3}\cdot\left(-\dfrac{2}{3}\right)\)
\(=\dfrac{11}{4}+\dfrac{4}{9}\)
\(=\dfrac{115}{36}\)
c) \(\sqrt{16}\cdot\sqrt{4}-\sqrt{25}+2\sqrt{49}\)
\(=4\cdot2-5+2\cdot7\)
\(=8-5+14\)
\(=17\)
d) \(\dfrac{1}{\sqrt{36}}+\dfrac{\sqrt{25}}{6}-\sqrt{0,81}\)
\(=\dfrac{1}{6}+\dfrac{5}{6}-0,9\)
\(=1-0,9\)
\(=0,1\)
e) \(-\dfrac{\sqrt{9}}{16}+\dfrac{5}{\sqrt{36}}\)
\(=-\dfrac{3}{16}+\dfrac{5}{6}\)
\(=\dfrac{31}{48}\)
f) \(\dfrac{\sqrt{9}}{8}\cdot\dfrac{16}{\sqrt{225}}-\dfrac{3}{4\sqrt{4}}\cdot\dfrac{2}{5\cdot\sqrt{3^2}}\)
\(=\dfrac{3}{8}\cdot\dfrac{16}{15}-\dfrac{3}{4\cdot2}\cdot\dfrac{2}{5\cdot3}\)
\(=\dfrac{3\cdot16}{8\cdot15}-\dfrac{3\cdot2}{4\cdot2\cdot5\cdot3}\)
\(=\dfrac{2}{5}-\dfrac{1}{20}\)
\(=\dfrac{7}{20}\)