a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)

\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)

\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)

\(\Leftrightarrow12x-10-8+2x=0\)

\(\Leftrightarrow10x-18=0\)

\(\Leftrightarrow10x=18\)

hay \(x=\frac{9}{5}\)

Vậy: \(x=\frac{9}{5}\)

b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)

\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)

\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)

\(\Leftrightarrow42-9x-2x+14=0\)

\(\Leftrightarrow56-11x=0\)

\(\Leftrightarrow11x=56\)

hay \(x=\frac{56}{11}\)

Vậy: \(x=\frac{56}{11}\)

c) ĐKXĐ: x∉{3;-3}

Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)

\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow6-x+2x-6=-5x-15\)

\(\Leftrightarrow x+5x+15=0\)

\(\Leftrightarrow6x=-15\)

hay \(x=\frac{-5}{2}\)(tm)

Vậy: \(x=\frac{-5}{2}\)

d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)

e) ĐKXĐ: x∉{4;-4}

Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)

\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)

\(\Leftrightarrow8x+10-4x+16=0\)

\(\Leftrightarrow4x+26=0\)

\(\Leftrightarrow4x=-26\)

hay \(x=\frac{-13}{2}\)(tm)

Vậy: \(x=\frac{-13}{2}\)

a) (x + 5)² - (x - 3)² = 2x - 7

(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7

8(2x + 2)= 2x - 7

16x + 16 = 2x - 7

16x - 2x = - 7 - 16

14x = - 23

x = - 23/14

b) (2x - 3)(4x² + 6x + 9) = 98

(2x)³ - 3³ = 98

8x³ - 27 = 98

8x³ = 125

x³ = 125/8

x³ = (5/2)³

x = 5/2

a: \(=\dfrac{2^{36}}{2^{12}}=2^{24}\)

b: \(=3^{18}:\dfrac{3^2}{5^2}=3^{16}\cdot5^2\)

c: \(=-\dfrac{\left(a-b\right)^5}{\left(a-b\right)^3}=-\left(a-b\right)^2\)

d: \(\dfrac{\left(a-b\right)^7}{\left(b-a\right)^4}=\dfrac{\left(a-b\right)^7}{\left(a-b\right)^4}=\left(a-b\right)^3\)

\(a.4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\\ 4x+8-14x+7+27x-36=30\\ 17x+15=66\\ 17x=51\Rightarrow x=3\)

\(b.2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ =10x-16-12x+15=12x-16+11\\ -2x-1=12x-5\\ \Leftrightarrow-2x-12x=1-5\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{7}{2}\)

\(c.4x^2+3\left(2x^2+1\right)=2x\left(5x-7\right)\\ 4x^2+6x^2+3=10x^2-14x\\ 10x^2+3=10x^2-14x\\ \Leftrightarrow3=14x\\\Rightarrow x=\dfrac{3}{14}\)

\(d.x\left(x^2-7\right)=2x\left(\dfrac{1}{2}x^2+6\right)+8\\ x^3-7x=x^3+12x+8\\ \Leftrightarrow-7x=12x+8\\ \Leftrightarrow-7x-12x=8\\ \Leftrightarrow-19x=8\Rightarrow x=-\dfrac{8}{19}\)

a)

\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\) chia hết cho 55 (đpcm )

b)

\(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.33\) chia hết cho 33 (đpcm )

c)

\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)

\(=3^{22}\left(3^6-3^5-3^4\right)=3^{22}.405\) chia hết cho 405 (đpcm )

Để ý rằng tất cả các biểu thức 2 vế của 4 bài đều không âm, cho nên ta bình phương 2 vế:

a/

\(\left(x^2-x+7\right)^2=\left(-5x+1\right)^2\)

\(\Leftrightarrow\left(x^2-x+7\right)^2-\left(-5x+1\right)^2=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2+4x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+8=0\\x^2+4x+6=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

b/

\(\left(x^2+9\right)^2=\left(-6x+1\right)^2\)

\(\Leftrightarrow\left(x^2+9\right)^2-\left(-6x+1\right)^2=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x^2+6x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+10=0\left(vn\right)\\x^2+6x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

c/

\(\left(x^2+5x+7\right)^2-\left(3x+5\right)^2=0\)

\(\Leftrightarrow\left(x^2+2x+2\right)\left(x^2+8x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+2=0\left(vn\right)\\x^2+8x+12=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)

d/

\(\left(x^2+6x+9\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(x^2+4x+6\right)\left(x^2+8x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+6=0\left(vn\right)\\x^2+8x+12=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)

Câu 1: A

Câu 2: D

Câu 3: C

Câu 4: B

Câu 5: D

Câu 6: C

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