\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

1. Ta cân bằng 2 phương trình nhỏ:

8Al + 30HNO3 →8Al(NO3)3 + 3N2O + 15H2O (1)

Al +4 HNO3 →Al(NO3)3 + NO + 2H2O (2)

Do tỉ lệ thể tích N2O : NO = 1:3. Mà hệ số của N2O ở (1) là 3 do đó ta nhân cả 2 vế của (2) với 9 rồi cộng với (1)

8Al + 30HNO3 →8Al(NO3)3 + 3N2O + 15H2O (1)

9x    (Al + 4 HNO3 →Al(NO3)3 + NO + 2H2O) (2)

Cộng lại được:

17Al + 66HNO3 →17Al(NO3)3 + 3N2O + 9NO + 33H2O

4.

29x | Al→Al³⁺+3e

3x | 7N⁺⁵+ 29e→3N⁺² + 2N₂

Cân bằng: 29

5.

19x | Al→Al³⁺+3e

3x | 5N⁺⁵+ 19e→N⁺² + 2N₂⁺¹

Cân bằng: 19

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

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Để hệ pt có nghiệm duy nhât \(\dfrac{a+1}{1}\ne\dfrac{-1}{a-1}\Leftrightarrow a^2-1\ne-1\Leftrightarrow a^2\ne0\Leftrightarrow a\ne0\)

\(\left\{{}\begin{matrix}\left(a^2-1\right)x-\left(a-1\right)y=a^2-1\\x+\left(a-1\right)y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2x=a^2+1\\y=\dfrac{2-x}{a-1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{a^2+1}{a^2}\\y=\dfrac{2-\dfrac{a^2+1}{a^2}}{a-1}=\dfrac{\dfrac{a^2-1}{a^2}}{a-1}=\dfrac{\left(a^2-1\right)\left(a-1\right)}{a^2}\end{matrix}\right.\)

Ta có \(\dfrac{a^2+1}{a^2}-\dfrac{\left(a^2-1\right)\left(a-1\right)}{a^2}=0\)

\(\Leftrightarrow a^2+1-a^3+a^2+a-1=0\)

\(\Leftrightarrow-a^3+2a^2+a=0\Leftrightarrow a^2-2a-1=0\)

\(\Leftrightarrow\left(a-1\right)^2-2=0\Leftrightarrow\left(a-1-\sqrt{2}\right)\left(a-1+\sqrt{2}\right)=0\Leftrightarrow a=1\pm\sqrt{2}\)

PTHH: C2H2 + 2Br2 -> C2H2Br4

Mol: 0,1 <--- 0,2

nCH4 = 3,36/22,4 = 0,15 (mol)

%VCH4 = 0,15/(0,1 + 0,15) = 60%

%VC2H2 = 100% - 60% = 40%

a) `6/30 = (1 xx 2)/(5 xx 2) = 2/10   (10 : 5 = 2)`

b) `72/800 = (72 : 8)/(800:8) = 9/100   (800 : 100 = 8)`

c) `81/270 = (81 : 27)/(270 : 27) = 3/10   (270 : 10 = 27)`

d) `19/200 = 95/1000   (19/200 = (19 xx 5)/(200 xx 5) = 95/1000) (1000 : 200 = 5)`.