a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

kết quả cuối cùng =1/2

/ phần nha bạn

nếu đúng thì tick cho mk nha

ngại làm wá bạn thông cảm

a,\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)=\(\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{6^9.2^{10}+6^{10}.2^{10}}\)

=\(\dfrac{2^{19}.3^9+2^{18}.3^9.5}{6^9.2^{10}.\left(1+6\right)}\)=\(\dfrac{2^{18}.3^9.\left(2+5\right)}{6^9.2^{10}.7}\)=\(\dfrac{2^{18}.3^9}{6^9.2^{10}}=\dfrac{2^{10}.2^8.3^9}{2^9.3^9.2^{10}}=\dfrac{2^8}{2^8.2}=\dfrac{1}{2}\)

b, \(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{16}}{\dfrac{-23}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-29}{16}}{\dfrac{-29}{16}}=1\)

Bài 1:

\(a,\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{2^9.3^9.2^{10}+3^{10}.\left(2^2\right)^{10}}\)

\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.\left(3^8.3\right)}{2^9.2^{10}.3^9+3^{10}.2^{20}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^{19}.3^9+3^{10}.2^{20}}\)

\(=\dfrac{1+5.1.1}{1+3.2^2}=\dfrac{6}{13}\)

=\(\dfrac{2^{19}.27^3-15.\left(-4\right)^9.9^4}{6^9.2^{10}+\left(-12\right)^{10}}\)

=\(\dfrac{2^{19}.3^9-2^{18}.3^9.5}{6^9.2^{10}+6^{10}.2^{10}}\)

=\(\dfrac{2^{18}.3^9\left(2-2.5\right)}{6^9.2^{10}\left(1+6\right)}\)

=\(\dfrac{2^{18}.3^9.\left(-8\right)}{3^9.2^{19}.7}\)

=\(\dfrac{-8}{14}=\dfrac{-4}{7}\)

Bài 1. ĐKXĐ thêm x ≠ 1 nữa ạ

1) Với x = 9 tmđk, thay vào A ta được : \(A=\dfrac{2\sqrt{9}+1}{9^2}=\dfrac{7}{81}\)

2) \(B=\left[\dfrac{4x}{\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}=\dfrac{4x-1}{x^2}\)

3) Để B < A thì \(\dfrac{4x-1}{x^2}< \dfrac{2\sqrt{x}+1}{x^2}\)

<=> \(\dfrac{4x-1}{x^2}-\dfrac{2\sqrt{x}+1}{x^2}< 0\)

<=> \(\dfrac{4x-2\sqrt{x}-2}{x^2}< 0\)

Vì x² > 0 ∀ x

=> \(4x-2\sqrt{x}-2< 0\)

<=> \(2x-\sqrt{x}-1< 0\)

<=> \(\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)< 0\)

Vì \(2\sqrt{x}+1\ge1>0\forall x\ge0\)

=> \(\sqrt{x}-1< 0\)<=> x < 1

Vậy với x < 1 thì B < A

Câu 3 : 

\(\left\{{}\begin{matrix}x-2y+\dfrac{1}{2x+3y}=2\\2x-4y+\dfrac{3}{2x+3y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y+\dfrac{1}{2x+3y}=2\\2\left(x-2y\right)+\dfrac{3}{2x+3y}=3\end{matrix}\right.\)

Đặt \(x-2y=t;\dfrac{1}{2x+3y}=z\)

Hệ phương trình tương đương 

\(\left\{{}\begin{matrix}t+z=2\\2t+3z=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=2-z\left(1\right)\\2t+3z=3\left(2\right)\end{matrix}\right.\)

Thế (1) vào (2) ta được : \(2\left(2-z\right)+3z=3\Leftrightarrow4-2z+3z=3\Leftrightarrow z=-1\)

\(\Rightarrow t=2-z=3\)

hay \(\left\{{}\begin{matrix}x-2y=3\\\dfrac{1}{2x+3y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\left(3\right)\\\dfrac{1}{2x+3y}=-1\left(4\right)\end{matrix}\right.\)

Thế (3) vào (4) ta được : \(\dfrac{1}{2\left(3+2y\right)+3y}=-1\Leftrightarrow\dfrac{1}{6+7y}=-1\Rightarrow-6-7y=1\Leftrightarrow-7y=7\Leftrightarrow y=-1\)

\(\Rightarrow x=3-2=1\)

Vậy \(\left(x;y\right)=\left(1;-1\right)\)

à câu trước em xin lỗi :( thiếu 

3) Kết hợp với ĐKXĐ => Với \(0\le x< 1\)thì B < A

Câu III

2) a) Ta có : Δ = b² - 4ac

= [ -(m-3) ]² - 4( 2m - 11 )

= m² - 6m + 9 - 8m + 44

= m² - 14m + 53 = ( m - 7 )² + 4 ≥ 4 > 0 ∀ m

hay pt luôn có hai nghiệm phân biệt với mọi m (đpcm)

b) Theo hệ thức Viète ta có : \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m-3\\x_1x_2=\dfrac{c}{a}=2m-11\end{matrix}\right.\)

Theo định lí Pythagoras ta có :

(CGV₁)² + (CGV₂)² = CH²

<=> x₁² + x₂² = 4²

<=> ( x₁ + x₂ )² - 2x₁x₂ - 16 = 0

<=> ( m - 3 )² - 2( 2m - 11 ) - 16 = 0

<=> m² - 6m + 9 - 4m + 22 - 16 = 0

<=> m² - 10m + 15 = 0 

Δ' = b'² - ac = 25 - 15 = 0

Δ' > 0, áp dụng công thức nghiệm => m = 5 ± √10

Vậy với m = 5 ± √10 thì thỏa mãn đề bài

a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

1. ĐKXĐ: $x>0; x\neq 9$

\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

3. ĐKXĐ: $a\geq 0; a\neq 1$

\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)

\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)