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Bài 1:
\(\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y-z\right)^2\)
\(=x^2\)
Bài 2:
đk: \(x\ne\left\{0;-1;-2;-3;-4;-5\right\}\)
Xét BT trái ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}\)
\(=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}\)
GT của biểu thức lớn sẽ là: \(\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1\) không phụ thuộc vào biến
=> đpcm
Bài 1.
( x - y + z ) + ( z - y )2 + ( x - y + z )( 2y - 2z )
= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )2
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
Bài 2. ĐKXĐ tự ghi nhé :))
\(\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)\)
\(=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1\)
=> đpcm
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
\(a,\frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}\)
\(\frac{13}{12}x=\frac{13}{12}\Rightarrow x=1\)
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow27x-2x-4x-27+2=0\)
\(\Leftrightarrow21x=25\)
\(\Leftrightarrow x=\frac{25}{21}\)
Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !
\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)
\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)
\(\Leftrightarrow-20x-12=56\)
\(\Leftrightarrow-20x=68\)
\(\Leftrightarrow x=-\frac{17}{5}\)
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