\(X^2=49\\ Mà:7^2=49;\left(-7\right)^2=49\\ \Rightarrow X=7.hoặc.x=-7\\ ----\\ b,\left(5x+1\right)^2=121=11^2=\left(-11\right)^2\\ Nên:5x+1=11.hoặc.5x+1=-11\\ Nên:5x=10.hoặc.5x=-12\\ Vậy:x=2.hoặc.x=-\dfrac{12}{5}\\ ---\\ 3x+36=-7x-64\\ \Rightarrow3x+7x=-64-36\\ \Rightarrow10x=-100\\ \Rightarrow x=-\dfrac{100}{10}=-10\\ ---\\ -5x-1178=14x+145\\ \Rightarrow14x+5x=-1178-145\\ \Rightarrow19x=-1323\\ \Rightarrow x=\dfrac{-1323}{19}\)

\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)

\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)

a: Thay x=49 vào A, ta được:

\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)

b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)

P>1 khi P-1>0

=>\(\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}>0\)

=>\(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)

a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

b) \(\Rightarrow x^2-13x+22=0\)

\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)

c) \(\Rightarrow x^2-3x-10=0\)

\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, Ta có : \(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\)

=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)

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

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

=> \(\left(x-1\right)\left(4x-3\right)=0\)

=> \(\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{1,\frac{3}{4}\right\}\)

b, Ta có : \(5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\)

=> \(5x^2+15x-27x-27=x^2-36\)

=> \(5x^2+15x-27x-27-x^2+36=0\)

=> \(4x^2-12x+9=0\)

=> \(\left(2x-3\right)^2=0\)

=> \(x=\frac{3}{2}\)

Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2}\right\}\)

\(a.\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{1;\frac{3}{4}\right\}\)

\(b.5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27-x^2+36=0\\ \Leftrightarrow4x^2-12x+9=0\\ \Leftrightarrow\left(2x-3\right)^2=0\\ \Leftrightarrow x=\frac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\frac{3}{2}\right\}\)

Chúc bạn học tốt!!!!!!!!!!!

a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)

\(=4.5+14:7\)

\(=20+2=22\)

b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\)

\(=36:\sqrt{18^2}-13\)

\(=36:18-13\)

\(=2-13=-11\)

c) \(\sqrt{\sqrt{81}}\)                   d) \(\sqrt{3^2}+4^2\)

\(=\sqrt{\sqrt{9^{^2}}}\)                   \(=3+16\)

\(=\sqrt{9}\)                          \(=19\)

a: \(\Leftrightarrow x^2-2x+1-x^2-2x-1=2x-6\)

=>2x-6=-4x

=>6x=6

hay x=1

b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

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

=>(x-3)(-4x+1)=0

=>x=3 hoặc x=1/4

c: \(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)-x^2+4x-4=0\)

\(\Leftrightarrow3x^2+16x+5-3x^2+48=0\)

=>16x+53=0

hay x=-53/16

d: \(\Leftrightarrow x^3+4x^2-9x-36=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-9\right)=0\)

hay \(x\in\left\{-4;3;-3\right\}\)

b)x^2-9=(x-3)(5x+2)

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

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

\(\Leftrightarrow\left(x-3\right)\left(1-4x\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\1-4x=0\end{matrix}\right.\left\{{}\begin{matrix}x=0+3\\x=1:4\end{matrix}\right.\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(a,\left(x-1\right)^2-\left(x+1\right)^2=2\left(x-3\right)\\ \Leftrightarrow x^2-2x+1-x^2-2x-1=2x-6\\ \Leftrightarrow-4x-2x=-6\\ \Leftrightarrow-6x=-6\\ \Leftrightarrow x=1\)

\(b,x^2-9=\left(x-3\right)\left(5x+2\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(c,\left(2x+3\right)^2-3\left(x-4\right)\left(x+4\right)=\left(x-2\right)^2\\ \Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\\ \Leftrightarrow4x^2+12x+9-3x^2+48-x^2+4x-4=0\\ \Leftrightarrow16x+53=0\\ \Leftrightarrow x=\dfrac{-53}{16}\)

\(d,x^3+4x^2-9x-36=0\\ \Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35