22: \(x+2\sqrt{x-1}=\left(\sqrt{x-1}+1\right)^2\)

24: \(-6x+5\sqrt{x}+1=\left(\sqrt{x}-1\right)\left(-6\sqrt{x}-1\right)\)

21: \(x^2-3x\sqrt{y}+2y\)

\(=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)

\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)

\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)

23: \(\sqrt{x^3}-2\sqrt{x}-x\)

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

\(=\sqrt{x}\left(x-\sqrt{x}-2\right)\)

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

Lời giải:

Theo định lý Viet:

$x_1+x_2=3$

$x_1x_2=-7$

Khi đó:
$A=\frac{1}{x_1-1}+\frac{1}{x_2-1}=\frac{x_2-1+x_1-1}{(x_1-1)(x_2-1)}$

$=\frac{(x_1+x_2)-2}{x_1x_2-(x_1+x_2)+1}=\frac{3-2}{-7-3+1}=\frac{-1}{9}$

$E=x_1^4+x_2^4=(x_1^2+x_2)^2-2(x_1x_2)^2=[(x_1+x_2)^2-2x_1x_2]^2-2(x_1x_2)^2$
$=[3^2-2(-7)]^2-2(-7)^2=431$

a)

\(\left(x^2-1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left[\left(x+2\right)^2-1\right]=\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)\)

\(\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)

dặt x^2+2x-1=t(*)

(a) \(\Leftrightarrow\left(t-2\right)\left(t+2\right)=192\) \(\Leftrightarrow t^2-4=192\Rightarrow t^2=196\Rightarrow\left\{\begin{matrix}t=-14\\t=14\end{matrix}\right.\)

Thay t vào (*) => x (tự làm)

a) (x-1)(x+1)(x+1)(x+3)=192. \(\Leftrightarrow\) (x+1)²(x-1)(x+3)=192 \(\Leftrightarrow\) (x²+2x+1) (x²+2x-3)=192 Đặt x²+2x+1=t thì x²+2x-3=t-4 ta có t(t-4)=192 \(\Leftrightarrow\) t²-4t-192=0 \(\Leftrightarrow\) t=-12 hoặc t=16 Với t=-12 thì (x+1)²=-12 ( vô lí ) Với t=16 thì (x+1)²=16 \(\Leftrightarrow\) x=-5 hoặc x=3 b) x\(^5\)+x⁴-2x⁴-2x³+5x³+5x²-2x²-2x+x+1=0 \(\Leftrightarrow\) x⁴(x+1)-2x³(x+1)+5x²(x+1)-2x(x+1)+(x+1)=0 \(\Leftrightarrow\) (x+1)(x⁴-2x³+5x²-2x+1)=0 \(\Leftrightarrow\) x=-1 ( CM x⁴-2x³+5x²-2x+1 vô nghiệm ) c) x⁴-x³-2x³+2x²+2x²-2x-x+1=0 \(\Leftrightarrow\) x³(x-1)-2x²(x-1)+2x(x-1)-(x-1)=0 \(\Leftrightarrow\) (x-1)(x³-2x²+2x-1)=0 \(\Leftrightarrow\) (x-1)(x-1)(x²-x+1)=0 \(\Leftrightarrow\) x-1=0 ( vì x²-x+1=(x-\(\frac{1}{2}\))²+\(\frac{3}{4}\)>0 với mọi x) \(\Leftrightarrow\) x=1

Ở phần b chứng minh vô nghiệm là ( x\(^4\)-2x³+x²)+(3x²-3x+\(\frac{3}{4}\))+\(\frac{5}{4}\)=0 \(\Leftrightarrow\) (x²-x)²+3(x+\(\frac{1}{2}\))²+\(\frac{5}{4}\)=0 ( vô lí)

\(5\cdot8^x-2^{3x+1}=192\Leftrightarrow5.2^{3x}-2^{3x+1}=192\Leftrightarrow2^{3x}\cdot\left(5-2\right)=192\)

\(\Leftrightarrow2^{3x}\cdot3=192\Leftrightarrow2^{3x}=\frac{192}{3}=64=2^6\Leftrightarrow x=3\)

Vậy x=3.

Chúc bạn học tốt!

5.8^x-2^3x+1=192

5.(2³)^x-2^3x+1=192

5.2^3x-2^3x+1=192

2^3x.(5-2)=192

2^3x.3=192

2^3x=192/3=64

2^3x=2⁶

suy ra 3x=6; x=2

Vậy x=2

\(5.8^x-2^{3x+1}=192\)

<= \(5.2^{3x}-2^{3x}.2=192\)

<=> \(2^{3x}.\left(5-2\right)=192\)

<=> \(2^{3x}=64\)

<=> \(2^{3x}=2^6\)

<=> \(3x=6\)

<=> \(x=2\)

học tốt

(12 - 3x + 2) x 4 =192

(12-3x+2) = 768

=> (12-3x)=768 ??????

Đến đây hơi vô lí xem lại hộ đề nhá @Trần Ngọc Minh Thư

(12-3x+2).4=192

<=>14-3x=48

<=>3x=-34

<=>x=-34/3

\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)

\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)

\(\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5+\left(-\dfrac{1}{4}\right)\)

\(=\left(-0,75\right)-\left(-1-\dfrac{2}{3}\right)\cdot\dfrac{1}{2}-0,25\)

\(=\left(-0,75-0,25\right)+\dfrac{5}{6}\)

\(=-1+\dfrac{5}{6}\)

\(=-\dfrac{11}{6}\)

_________________

\(\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)

\(=\left(-\dfrac{9}{6}+\dfrac{4}{6}\right)^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)

\(=\left(\dfrac{-5}{6}\right)^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)

\(=\dfrac{25}{36}\cdot\dfrac{24}{25}-\dfrac{1}{5}\)

\(=\dfrac{2}{3}-\dfrac{1}{5}\)

\(=\dfrac{7}{15}\)

lười quá

\(a,\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(\dfrac{-3+2}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right)\times2-\dfrac{1}{4}\\ =-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{4}\\ =\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)+\dfrac{2}{3}\\ =-\dfrac{4}{4}+\dfrac{2}{3}\\ =-1+\dfrac{2}{3}\\ =\dfrac{-3+2}{3}=-\dfrac{1}{3}\)

\(b,\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(\dfrac{-3\times3+2\times2}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(-\dfrac{5}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{25}{36}\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{2}{3}-\dfrac{1}{5}\\ =\dfrac{2\times5-3}{15}=\dfrac{7}{15}\)