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âncsi hình sau mình nhìn muốn gãyy cổ rồi :v

dài v

a) x(x-y)+(x-y)=(x+1)(x-y)

b) 2x+2y -x(x+y)= 2(x+y)-x(x+y)=(2-x)(x+y)

\(a,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4=49\)

\(\Leftrightarrow12x=36\)

\(\Rightarrow x=3\)

b) \(16x^2-\left(4x-5\right)^2=15\)

\(\Rightarrow16x^2-16x^2+40x-25=15\)

\(\Rightarrow x=1\)

d) \(\left(2x+5\right)\left(8x-7\right)-\left(-4x-3\right)^2=16\)

\(\Leftrightarrow16x^2-14x+40x-35-16x^2+24x-9=16\)

\(\Leftrightarrow50x=60\)

\(\Rightarrow x=\dfrac{6}{5}\)

e) \(49x^2+12x+1=0\)

\(\Leftrightarrow7x+1=0\)

\(\Rightarrow x=\dfrac{-1}{7}\)

f) \(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2-2x+1+y^2+4x+5=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

23.27. \(x^2-y^2-2x+1\)

\(=\left(x-1\right)^2-y^2\)

\(=\left(x-1-y\right)\left(x-1+y\right)\)

23.25.

\(\left(x^2-4x\right)^2+\left(x-2\right)^2-10\)

\(=\left(x^2-4x\right)^2-4+\left(x-2\right)^2-6\)

\(=\left(x^2-4x+4\right)\left(x^2-4x-4\right)+x^2-4x+4-6\)

\(=\left(x^2-4x+4\right)\left(x^2-4x-10\right)\)

23.23

\(x^3-2x^2-6x+27\)

\(=\left(x^3+27\right)-2x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+9-2x\right)\)

\(=\left(x+3\right)\left(x^2-5x+9\right)\)

23.27

\(x^2-y^2-2x+1\)

\(=\left(x^2-2x+1\right)-y^2\)

\(=\left(x-1\right)^2-y^2\)

\(=\left(x-1-y\right)\left(x-1-y\right)\)

b)\(B=1^2-2^2+3^2-4^2+...-2016^2+2017^2\)

\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+...+\left(2015^2-2016^2\right)+2017^2\)

\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2015-2016\right)\left(2015+2016\right)+2017^2\)

\(=-1\cdot\left(1+2\right)+\left(-1\right)\cdot\left(3+4\right)+...+\left(-1\right)\cdot\left(2015+2016\right)+2017^2\)

\(=-1\cdot\left(1+2+...+2015+2016\right)+2017^2\)

\(=-1\cdot\dfrac{2016\cdot\left(2016+1\right)}{2}+2017^2\)

\(=-2033136+4068289=2035153\)

c)\(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(=2^{64}-1-2^{64}=-1\)

Câu 1:

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

Thay \(a+b+c=0\) vào biểu thức ta được:

\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow a^3+b^3=3abc\left(đpcm\right)\)

Vậy \(a^3+b^3=3abc\) khi \(a+b+c=0\)

Câu 3:

\(\text{a) }x^2+x+1\\ =x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{4}\right)^2\right]+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ \text{Ta có : }\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ \text{ Vậy biểu thức luôn nhận giá trị dương}\text{ }\forall x\\ \)

\(\text{b) }2x^2+2x+1\\ =2x^2+2x+\dfrac{1}{2}+\dfrac{1}{2}\\ =2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}\\ =2\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{1}{2}\\ =2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\\ \text{Ta có: }2\left(x+\dfrac{1}{2}\right)^2\forall x\\ 2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\forall x\\ \text{Vậy giá trị của biểu thức luôn nhận giá trị dương }\forall x\\ \)

39

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đề 1 bài 4

xét tam gics ABC và tam giác HBA có

góc B chung

góc BAC = góc BHA (=90 độ)

=> tam giác ABC đồng dạng vs tam giác HBA (g.g)

=> AB/HB=BC/AB=> AB^2=HB *BC

áp dụng đl py ta go trog tam giác vuông ABC có

BC^2 = AB^2 +AC^2=6^2+8^2=100

=> BC =\(\sqrt{100}\)=10 cm

ta có tam giác ABC đồng dạng vs tam giác HBA (cm câu a )

=> AC/AH=BC/BA=>AH=8*6/10=4.8CM

=>AB/BH=AC/AH=> BH=6*4.8/8=3,6cm

=>HC =BC-BH=10-3,6=6,4cm

dề 1 bài 1

5x+12=3x -14

<=>5x-3x=-14-12

<=>2x=-26

<=> x=-12

vạy S={-12}

(4x-2)*(3x+4)=0

<=>4x-2=0<=>x=1/2

<=>3x+4=0<=>x=-4/3

vậy S={1/2;-4/3}

đkxđ : x\(\ne2;x\ne-3\)

\(\dfrac{4}{x-2}+\dfrac{1}{x+3}=0\)

<=> 4(x+3)/(x-2)(x+3)+1(x-2)/(x-2)(x+3)

=> 4x+12+x-2=0

<=>5x=-10

<=>x=-2 (nhận)

vậy S={-2}