RÚT GỌN
a) \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
b)\(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
Rút gọn rồi tính giá trị biểu thức :
\(E=\frac{\left(a-x\right)^2}{a\left(b-a\right)\left(c-a\right)}+\frac{\left(b-x\right)^2}{b\left(a-b\right)\left(c-b\right)}+\frac{\left(c-x\right)^2}{c\left(a-c\right)\left(b-c\right)}\)
Biết : \(1-\frac{x^2}{abc}=0\)
Rút gọn các biểu thức sau:
A =\(\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a'^6+b'^6\right)\)
B = \(\left(x+1\right)^2+\left(x+2\right)^{^2}-2\left(x-3\right)^2-18\left(x-1\right)\)
C = \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(a+b\right)^2\)
D = \(\left(a+b-c\right)^2-\left(a-b+c\right)^2-4a\left(b-c\right)\)
Mk cần gấp! Mong mọi người giúp ạ!
Bài 1: Tìm x
a) \(\left(5-2x\right)^2-16=0\)
b) \(x^2-4x=29\)
c) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+9.\left(x+1\right)^2=15\)
d) \(2.\left(x-5\right).\left(x+5\right)-\left(x+2\right).\left(2x-3\right)+x.\left(x^2-8\right)=\left(x+1\right).\left(x^2-x+1\right)\)
Bài 2: Rút gọn
a) \(\left(x^2+x+1\right).\left(x^2-x+1\right).\left(x^2-1\right)\)
b) \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2.\left(b-c\right)^2\)
c) \(\left(a+b+c\right)^2-\left(a+b\right)^2-\left(a+c\right)^2-\left(b+c\right)^2\)
d) \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
1. a) $(5-2x)^2-16=0$
$=>(5-2x)^2-4^2=0$
$=>(5-2x-4)(5-2x+4)=0$
$=>(1-2x)(9-2x)=0$
\(=>\left[{}\begin{matrix}1-2x=0=>x=0,5\\9-2x=0=>x=4,5\end{matrix}\right.\)
b) $x^2-4x=29$
$=>x^2-4x-29=0$
$=>(x^2-4x+4)-33=0$
$=>(x-2)^2-(\sqrt{33})^2=0$
$=>(x-2-\sqrt{33})(x-2+\sqrt{33})=0$
\(=>\left[{}\begin{matrix}x-2-\sqrt{33}=0=>x=\sqrt{33}+2\\x-2+\sqrt{33}=0=>x=2-\sqrt{33}\end{matrix}\right.\)
Bài 1:
a) \(\left(5-2x\right)^2-16=0\) (1)
\(\Leftrightarrow\left(5-2x\right)^2=16\)
\(\Leftrightarrow5-2x=\pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{1}{2};\dfrac{9}{2}\right\}\)
b) \(x^2-4x=29\) (2)
\(\Leftrightarrow x^2-4x-29=0\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{33}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+2\sqrt{33}}{2}\\x=\dfrac{4-2\sqrt{33}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{2-\sqrt{33};2+\sqrt{33}\right\}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\) (3)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9x^2+18x+9=15\)
\(\Leftrightarrow x^3+27x-27-x^3+27+18x+9=15\)
\(\Leftrightarrow45x+9=15\)
\(\Leftrightarrow45x=15-9\)
\(\Leftrightarrow45x=6\)
\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{2}{15}\right\}\)
d) \(2\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(2x-3\right)+x\left(x^2+8\right)=\left(x+1\right)\left(x^2-x+1\right)\)(4)
\(\Leftrightarrow2\left(x^2-25\right)-\left(2x^2-3x+4x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-\left(2x^2+x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-2x^2-x+6-8x=1\)
\(\Leftrightarrow-44-9x=1\)
\(\Leftrightarrow-9x=1+45\)
\(\Leftrightarrow-9x=45\)
\(\Leftrightarrow x=-5\)
Vậy tập nghiệm phương trình (4) là \(S=\left\{-5\right\}\)
Bài 2:
a) \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x^4-x^3+x^2+x^3-x^2+x+x^2-x+1\right)\left(x^2-1\right)\)
\(=x^6-x^4+x^4-x^2+x^2-1\)
\(=x^6-1\)
b) \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=a^2+b^2+\left(-c\right)^2+2ab-2ac+a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc-2\left(b^2+2bc+c^2\right)\)
\(=a^2+b^2+\left(-c\right)^2-2bc+a^2+\left(-b\right)^2+c^2-2bc-2b^2-4bc-2c^2\)
\(=2a^2-b^2+c^2-8bc+\left(-b\right)^2-c^2\)
\(=2a^2-b^2-8bc+b^2\)
\(=2a^2-8bc\)
c) \(\left(a+b+c\right)^2-\left(a+b\right)^2-\left(a+c\right)^2-\left(b+c\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-\left(a^2+2ab+b^2\right)-\left(a^2+2ac+b^2\right)-\left(b^2+2bc+c^2\right)\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-a^2-2ab-b^2-a^2-2ac-c^2-b^2-2bc-c^2\)
\(=-a^2-b^2-c^2\)
d) \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2-2ab-2ac+2bc+\\ b^2+c^2+a^2-2bc-2ab+2ac+c^2+a^2+b^2-2ac-2bc+2ab\)
\(=2a^2+2b^2+2c^2+2\cdot\left(-b\right)^2+2\cdot\left(-c\right)^2+2\cdot\left(-a\right)^2\)
\(=2a^2+2b^2+2c^2+2b^2+2c^2+2a^2\)
\(=4a^2+4b^2+4c^2\)
Bài 1 : Rút gọn
a)\(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a+b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
Rút gọn
a) \(\left(x+2y\right)^2+4y\)
b) \(\left(2x-3\right)\left(2x+3\right)-4x^2\)
c)\(\left(3x+1\right)^2-\left(x+1\right)\left(x-1\right)\)
Lời giải:
a. Biểu thức này không có khả năng rút gọn. Khai triển ra cũng được nhưng không làm gọn được bạn nhé.
b. $=(2x)^2-3^2-4x^2=4x^2-9-4x^2=-9$
c. $=(3x)^2+2.3x+1^2-(x^2-1)=9x^2+6x+1-x^2+1=8x^2+6x+2$
Bài 1 : rút gọn các biểu thức sau
A = \(\left(3x+1\right)^2-2\left(3x+1\right)\left(5x+5\right)+\left(5x+5\right)^2\)
B = \(\left(a+b+c\right)^2\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-b-a\right)^2\)
C = \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
Bài 2 : chứng minh các biểu thức sau không phụ thuộc vào biến x và y
A = \(\left(2x-1\right)\left(x^2+x-1\right)-\left(x-5\right)^2-2\left(x+1\right)\left(x^2-x+1\right)-7\left(x-2\right)\)
Bài 1: Rút gọn
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a+b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
b)\(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Rút gọn biểu thức
a.\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
b.\(\frac{a^3-b^3+c^3+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
c.\(\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
d.\(\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\left(x^{16}-x^8+1\right)\)
MONG CÁC BẠN CÓ THỂ BỎ RA VÀI PHÚT ĐỂ GIÚP MÌNH=))NÓ CŨNG GIÚP BẠN ÔN TẬP ĐƯỢC CÁC BÀI CHUẨN BỊ CHO KÌ THÌ MÀ=))MÌNH XIN CẢM ƠN RẤT RẤT NHIỀU
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
Rút gọn các biểu thức sau :
a) \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3-x^3-3x\left(x+1\right)\left(x-1\right)\)
c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)
d) \(100^2-99^2+98^2+97^2+......+2^2-1^2\)
e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
f) \(\left(a+b+c\right)^{^{ }2}+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
a,b,c,f tìm cách áp dụng HĐT vào nhé! động não tí xem :)
d) Sửa đề :\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=199+195+...+3\)
Khi đó tổng sẽ là:
\(\dfrac{\left(199+3\right)\left[\dfrac{\left(199-3\right)}{4}+1\right]}{2}=5050.\)
e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1\)
\(=2^{128}.\)