1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)

B: rút gọn

a) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-6x^2+12x\)

\(=x^3-6x^2+12x-8\)

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

b) Ta có: \(\left(2x+5\right)\left(5-2x\right)+\left(x-5\right)\left(4x+5\right)\)

\(=25-4x^2+4x^2+5x-20x-25\)

=-15x

Nếu ol thì tham khảo nah nguoiemtinhthong.

1.1

2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1

⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)

Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0

pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0

a=2ba=2b v a=13ba=13b

Các bạn tự giải quyết tiếp nhé.

1.2

TXĐ D=[1;+∞)D=[1;+∞)

đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0

pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0

⇔a=b⇔a=b v a=23ba=23b

...

1.3

D=[3;+∞)D=[3;+∞)

Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0

pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2

⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0

⇒a=5b⇒a=5b
...

1.4

ĐK

⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)

⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)

Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)

⇔2a2+2b2=3ab

1.5

Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)

⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x

⇔t2−t−4x2+2x=0t2−t−4x2+2x=0

Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2

⇒t=1−2xt=1−2x hoặc t=2xt=2x

1.1

2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1

2(.2+x+1)+3(x-1)

3a+b=11a2-19b2

tóm tắt

\(a.\) \(x^3-25x=0\)

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

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

TH1: \(x=0\)

TH2: \(x+5=0\Rightarrow x=-5\)

TH3: \(x-5=0\Rightarrow x=5\)

a, x³-25x = 0

\(\Leftrightarrow\) x( x²- 25) = 0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: S= { 0; 5; -5}

b, (2x+3)² = (x+4)²

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

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=\dfrac{-7}{3}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm: S= {1; \(\dfrac{-7}{3}\)}

c, (2x-1)² - (2x-5)(2x+5) = 18

\(\Leftrightarrow\) 4x²- 4x+ 1 - ( 4x²- 25) = 18

\(\Leftrightarrow\) 4x²- 4x+ 1- 4x²+ 25 = 18

\(\Leftrightarrow\) -4x + 26 = 18

\(\Leftrightarrow\) -4x = -8

\(\Leftrightarrow\) x = 2

Vậy phương trình có tập nghiệm S = { 2}

d, x³ - 8 = ( x-2)³

^{\(\Leftrightarrow\)} x³ - 8 = x³ - 6x² + 12x -8

\(\Leftrightarrow\) 6x² - 12x = 0

\(\Leftrightarrow\) 6x( x- 2) = 0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm: S = {0; 2}

1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)

\(=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)

\(=\frac{x^3+2x+2x^2-50+50-5x}{2x\left(x+5\right)}\)

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

\(=\frac{\left(x-1\right)\left(x+3\right)}{2\left(x+5\right)}\)

a. \(3x^2-2x\left(5+1.5x\right)+10\)

\(=3x^2-10x-3x^2+10\)

\(=-10x+10\)

b. \(\left(x^2-2x+3\right)\left(x-4\right)\)

\(=x^3-2x^2+3x-4x^2+8x-12\)

\(=x^3-6x^2+11x-12\)

c. \(\left(5x+2\right)\left(2x^2-3x-1\right)\)

\(=10x^3-15x^2-5x+4x^2-6x-2\)

\(=10x^3-11x^2-11x-2\)

d. \(\left(25x^2+10xy+4y^2\right)\left(5x+2y\right)\)

\(=125x^3+50x^2y+20xy^2+50x^2y+10xy^2+6y^3\)

\(=125x^3+100x^2y+30xy^2+6y^3\)

e. \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)

\(=20x^5-4x^4+2x-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-1\)

\(=20x^5-9x^4+9x-16x^2+11x^3+1\)

c) x² + 9x = 10

x² + 9x - 10 = 0

=> x² - x + 10x - 10 = 0

=> x(x - 1) + 10(x - 1) = 0

=> (x + 10)(x - 1) = 0

=> \(\orbr{\begin{cases}x=-10\\x=1\end{cases}}\)

d) 2x² + 9x = 35

=> 2x² + 9x - 35 = 0

=> 2x² + 14x - 5x - 35 = 0

=> 2x(x + 7) - 5(x + 7) = 0

=> (x + 7)(2x - 5) = 0

=> \(\orbr{\begin{cases}x=-7\\x=\frac{5}{3}\end{cases}}\)

(x² - 2x - 1)² - 5(x² - 2x - 1) - 14 = 0

=> (x² - 2x - 1)² + 2(x² - 2x - 1) - 7(x² - 2x - 1) - 14 = 0

=> (x² - 2x - 1)(x² - 2x + 1) - 7(x² - 2x + 1) = 0

=> (x² - 2x + 1)(x² - 2x - 8) = 0

=> (x - 1)² (x - 4)(x + 2) = 0

=> x = 1 hoặc x = 4 hoặc x = -2

e) (2k² + 5k + 1)² - 12(2k² + 5k + 1) + 32 = 0

=> (2k² + 5x + 1)² - 4(2k² + 5k + 1) - 8(2k² + 5k + 1) + 32 = 0

=> (2k² + 5k + 1)(2k² + 5k - 3) - 8(2k² + 5k - 3) = 0

=> (2k² + 5k - 3)(2k² + 5k - 7) = 0

=> (2k² + 6k - k - 3)(2k² - 2x + 7k - 7) = 0

=> (k + 3)(2k - 1)(k - 1)(2k + 7) = 0

=> k = -3 hoặc k = 1/2 hoặc k = 1 hoặc k = -7/2

1.x² + 6x = 0 < như này nhỉ ? >

⇔ x( x + 6 ) = 0

⇔ x = 0 hoặc x + 6 = 0

⇔ x = 0 hoặc x = -6

2. x² - 25x + 250 = 0

⇔ ( x² - 25x + 625/4 ) + 375/4 = 0

⇔ ( x - 25/2 )² = -375/4 ( vô lí )

=> Phương trình vô nghiệm

3. x² + 9x = 10

⇔ x² + 9x - 10 = 0

⇔ x² - x + 10x - 10 = 0

⇔ x( x - 1 ) + 10( x - 1 ) = 0

⇔ ( x - 1 )( x + 10 ) = 0

⇔ x - 1 = 0 hoặc x + 10 = 0

⇔ x = 1 hoặc x = -10

4. 2x² + 9x = 35

⇔ 2x² + 9x - 35 = 0

⇔ 2x² + 14x - 5x - 35 = 0

⇔ 2x( x + 7 ) - 5( x + 7 ) = 0

⇔ ( x + 7 )( 2x - 5 ) = 0

⇔ x + 7 = 0 hoặc 2x - 5 = 0

⇔ x = -7 hoặc x = 5/2

5. ( x² - 2x - 1 )² - 5( x² - 2x - 1 ) - 14 = 0

Đặt t = x² - 2x - 1

bthuc ⇔ t² - 5t - 14 = 0

          ⇔ t² - 7t + 2t - 14 = 0

          ⇔ t( t - 7 ) + 2( t - 7 ) = 0

          ⇔ ( t - 7 )( t + 2 ) = 0

          ⇔ ( x² - 2x - 1 - 7 )( x² - 2x - 1 + 2 ) = 0

          ⇔ ( x² - 4x + 2x - 8 )( x - 1 )² = 0

          ⇔ ( x - 4 )( x + 2 )( x - 1 )² = 0

          ⇔ x - 4 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0

          ⇔ x = 4 hoặc x = -2 hoặc x = 1

6. ( 2k² + 5k + 1 )² - 12( 2k² + 5k + 1 ) + 32 = 0

Đặt t = 2k² + 5k + 1

bthuc ⇔ t² - 12t + 32 = 0

          ⇔ t² - 8t - 4t + 32 = 0

          ⇔ t( t - 8 ) - 4( t - 8 ) = 0

          ⇔ ( t - 8 )( t - 4 ) = 0

          ⇔ ( 2k² + 5k + 1 - 8 )( 2k² + 5k + 1 - 4 ) = 0

          ⇔ ( 2k² - 2k + 7k - 7 )( 2k² - k + 6k - 3 ) = 0

          ⇔ ( k - 1 )( 2k + 7 )( 2k - 1 )( k + 3 ) = 0

          ⇔ k = 1 hoặc k = -7/2 hoặc k = 1/2 hoặc k = -3