Bài 1

1.(x-3)(x+2)-x(x-7)=15

\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)

\(\Leftrightarrow-6+6x=15\)

\(\Leftrightarrow6x=15+6\) =21

\(\Rightarrow x=\dfrac{21}{6}=3,5\)

2.(x-5)(x+5)+x(3-x)=20

\(\Leftrightarrow x^2-25+3x-x^2=20\)

\(\Leftrightarrow-25+3x=20\)

\(\Leftrightarrow3x=20+25=45\)

\(\Rightarrow x=\dfrac{45}{3}=15\)

3.(x-7)²-x(2+x)=-7

\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)

\(\Leftrightarrow-16x+49=-7\)

\(\Leftrightarrow-16x=-7-49=-56\)

\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)

Tiếp bài 1

4.(x-4)²-(x+4)(x-4)=-16

\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)

\(\Leftrightarrow-8x=-16\)

\(\Rightarrow x=\dfrac{-16}{-8}=2\)

5.(x-5)(x+5)-x(2-3x)=4x²-7

\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)

\(\Leftrightarrow-2x=18\)

\(\Rightarrow x=\dfrac{18}{-2}=-9\)

\(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-4\right)^2=8\left(x-3\right)\left(x+3\right)\)3)

\(\Leftrightarrow x^3+4^3-x\left(x-4\right)^2=8\left(x^2-3^2\right)\)

\(\Leftrightarrow x^3+64-x\left(x^2-8x+16\right)=8x^2-72\)

\(\Leftrightarrow x^3+64-x^3+8x^2-16x-8x^2-72=0\)

\(\Leftrightarrow-16x-8=0\)

\(\Leftrightarrow-8\left(2x-1\right)=0 \)

\(\Rightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy   \(x=\frac{1}{2}\)

$\textbf{1)}$

$(4x-1)^2-2(4x-1)(3x-7)+(7-3x)^2$

$=(4x-1)^2-2(4x-1)(3x-7)+(3x-7)^2$

$=\left[(4x-1)-(3x-7)\right]^2$

$=(x+6)^2.$

Tại $x=44$: $(44+6)^2=50^2=2500.$

$\textbf{2)}$

$(2x-5)^2-2(2x-5)(3x-4)+(4-3x)^2$

$=(2x-5)^2-2(2x-5)(3x-4)+(3x-4)^2$

$=\left[(2x-5)-(3x-4)\right]^2$

$=(x+1)^2.$

Tại $x=24$: $(24+1)^2=625.$

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

$\textbf{5)}$

$(4x-5)(x+2)-(x+5)(x-3)-3x^2-x$

$=(4x^2+3x-10)-(x^2+2x-15)-3x^2-x$

$=4x^2+3x-10-x^2-2x+15-3x^2-x$

$=5.$

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

=>1+3x-6=3-x

=>3x-5=3-x

=>4x=8

hay x=2(loại)

b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)

=>8-x-8x+56=-26

=>-9x+64=-26

=>-9x=-90

hay x=10(nhận)

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

\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)

\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)

=>-7x+10x=12-5

=>3x=7

hay x=7/3(nhận)

3: \(-\dfrac{1}{2}+\left|\dfrac{1}{3}x+\dfrac{2}{5}\right|=\dfrac{5}{7}\cdot\dfrac{3}{11}+\dfrac{8}{11}\cdot\dfrac{5}{7}=\dfrac{5}{7}\)

=>|1/3x+2/5|=5/7+1/2=17/14

=>1/3x+2/5=17/14 hoặc 1/3x+2/5=-17/14

=>x=171/70 hoặc x=-339/70